Differential Equations Class 12 Maths Important Questions Chapter 9

Differential Equation Exercise 9.1

Determine the Order And Degree (If Defined) Of Differential Equations

Question 1. \(\frac{d^4 y}{d x^4}+\sin \left(y^{\prime \prime \prime}\right)=0\)
Solution:

⇒ \(\frac{\mathrm{d}^4 \mathrm{y}}{\mathrm{dx}}+\sin \left(\mathrm{y}^{\prime \prime \prime}\right)=0 \Rightarrow \mathrm{y}^{\prime \prime \prime \prime}+\sin \left(\mathrm{y}^{\prime \prime \prime}\right)=0\)

The highest order derivative present in the differential equation is y””. Therefore, its order is four.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

Question 2. y’+5y = 0
Solution:

The given differential equation is: y’ + 5y = 0

The highest-order derivative present in the differential equation is y’. Therefore, its order is one.

It is a polynomial equation in y’. The highest power raised to y’ is 1. Hence, its degree is one.

Question 3. \(\left(\frac{\mathrm{ds}}{\mathrm{dt}}\right)^4+3 \mathrm{~s} \frac{\mathrm{d}^2 \mathrm{~s}}{\mathrm{dt}^2}=0\)
Solution:

The highest order derivative present in the given differential equation is \(\frac{\mathrm{d}^2 \mathrm{~s}}{\mathrm{dt}^2}=0\). Therefore, its order is two.

It is a polynomial equation in \(\frac{\mathrm{d}^2 \mathrm{~s}}{\mathrm{dt}^2}=0\) and \(\frac{ds}{dt}\). The power raised to \(\frac{\mathrm{d}^2 \mathrm{~s}}{\mathrm{dt}^2}=0\) is 1.

Hence, its degree is one.

Read and Learn More Class 12 Maths Chapter Wise with Solutions

Question 4. \(\left(\frac{d^2 y}{d x^2}\right)^2+\cos \left(\frac{d y}{d x}\right)=0\)
Solution:

The highest order derivative present in the given differential equation is \(\frac{d^2 y}{d x^2}\) order is 2.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

Question 5. \(\frac{d^2 y}{d x^2}=\cos 3 x+\sin 3 x\)
Solution:

⇒ \(\frac{d^2 y}{d x^2}=\cos 3 x+\sin 3 x \Rightarrow \frac{d^2 y}{d x^2}-\cos 3 x-\sin 3 x=0\)

The highest order derivative present in the differential equation is \(\frac{d^2 y}{d x^2}\). Therefore, its order is two.

It is a polynomial equation in \(\frac{d^2 y}{d x^2}\) and the power raised to \(\frac{d^2 y}{d x^2}\) is 1.

Hence, its degree is one.

Question 6. \(\left(y^{\prime \prime \prime}\right)^2+\left(y^{\prime \prime}\right)^3+\left(y^{\prime}\right)^4+y^3=0\)
Solution:

The highest order derivative present in the differential equation is \(\left(y^{\prime \prime \prime}\right)\).

Therefore, its order is three. The given differential equation is a polynomial equation \(iny^{\prime \prime \prime}, y^{\prime \prime} \text { and } y^{\prime} \text {. }\)

The highest power raised to \(y^{\prime \prime \prime}\) is 2. Hence, its degree is 2.

Question 7. \(y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0\)
Solution:

The highest order derivative present in the differential equation is \(y^{\prime \prime \prime}\). Therefore, its order is three.It is a polynomial equation in \(y^{\prime \prime \prime}\), y” and y’. The highest power raised to y is 1. Hence, its degree is 1.

Question 8. y’ + y = e^x
Solution:

y’ + y = e^x ⇒ y’ + y = e^x = 0

The highest-order derivative present in the differential equation is y’. Therefore, its order is one. The given differential equation is a polynomial equation and the highest power raised to y’ is one. Hence, its degree is one.

 

Quetsion 9. y”+(y’)² + 2y = 0
Solution:

The highest order derivative present in the differential equation is Therefore, its order is two.

The given differential equation is a polynomial equation in y” and y’ and the highest power raised to y” is one.

Hence, its degree is one. y”+ 2y’+ sin y = 0

Question 10. y” + 2y’ + sin y = 0
Solution:

The highest order derivative present in the differential equation is y”. Therefore, its order is two. This is a polynomial equation in y” and y’, and the highest power raised to y” is one.

Hence, its degree is one.

Question 11. The degree of the differential equation \(\left(\frac{d^2 y}{d x^2}\right)^3+\left(\frac{d y}{d x}\right)^2+\sin \left(\frac{d y}{d x}\right)+1=0\) is

1. 3
2. 2
3. 1
4. Not Defined

Solution:

The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.

Hence, the correct answer is D.

Question 12. The order of the differential equation \(2 x^2 \frac{d^2 y}{d x^2}-3 \frac{d y}{d x}+y=0\)

1. 2
2. 1
3. 0
4. Not Defined

Solution:

The highest order derivative present in the given differential equation is \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\). Therefore, its order is two.

Hence, the correct answer is A.

Differential Equation Exercise 9.2

Verify That The Given Function (Explicit Or Implicit) Is A Solution Of The Corresponding Differential Equation:

Question 1. y = e^x + 1 : y” – y’ = 0
Solution:

y = e^x + 1

Differentiating both sides of this equation with respect to x, we get: \(\frac{d y}{d x}=\frac{d}{d x}\left(e^x+1\right) \Rightarrow y^{\prime}=e^x\)….(1)

Now, again differentiating equation (1) with respect to x, we get: \(\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^x\right) \Rightarrow y^{\prime \prime}=e^x\)

Substituting the values of y’ and y” in the given differential equation, we get the L.H.S. as y” – y’ = e^x – e^x = 0 = R.H.S.

Thus, the given function is the solution of the corresponding differential equation.

Question 2. y = x² + 2x + C : y’ – 2x – 2 = 0
Solution:

y = x² + 2x + C

Differentiating both sides of this equation with respect to x, we get:

y’ = \(\frac{d}{dx}\)(X² + 2X + C) ⇒ y’ = 2x+2 dx

Substituting the value of y’ in the given differential equation, we get:

L.H.S. = y’ – 2x – 2 =2x + 2-2x-2 = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 3. y = cos x + C : y’ + sin x = 0
Solution:

y = cos x + C

Differentiating both sides of this equation with respect to x, we get:

y’ = \(\frac{d}{dx}\)(cos x + C) ⇒ y’ = -sin x

Substituting the value of y’ in the given differential equation, we get:

L.H.S. = y’ + sinx = – sinx + sinx = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 4. \(y=\sqrt{1+x^2}: y^{\prime}=\frac{x y}{1+x^2}\)
Solution:

y = \(\sqrt{1+x^2}\)

Differentiating both sides of the equation with respect to x, we get:

⇒ \(y^{\prime}=\frac{d}{d x}\left(\sqrt{1+x^2}\right) \Rightarrow y^{\prime}=\frac{1}{2 \sqrt{1+x^2}} \cdot \frac{d}{d x}\left(1+x^2\right)\)

⇒ \(y^{\prime}=\frac{2 x}{2 \sqrt{1+x^2}}\)

⇒ \(y^{\prime}=\frac{x}{\sqrt{1+x^2}} \Rightarrow y^{\prime}=\frac{x}{1+x^2} \times \sqrt{1+x^2}\)

⇒ \(y^{\prime}=\frac{x}{1+x^2} \cdot y \Rightarrow y^{\prime}=\frac{x y}{1+x^2}\)

∴ L.H.S. = R.H.S.

Hence, the given function is the solution of the corresponding differential solution.

Question 5. y = Ax : xy’ = y (x ≠ 0)
Solution:

y = Ax

Differentiating both sides of the equation with respect to x, we get: \(y^{\prime}=\frac{d}{d x}(A x) \Rightarrow y^{\prime}=A\)

Substituting the value of y’ in the given differential equation, we get:

L.H.S. = xy’ = x-A = Ax = y = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Question 6. \(y=x \sin x \quad: \quad x y^{\prime}=y+x \sqrt{x^2-y^2} \quad[x \neq 0 \text { and } x>y \text { or } x<-y]\)
Solution:

y = x sin x…..(1)

Differentiating both sides of this equation with respect to x, we get: \(y^{\prime}=\frac{d}{d x}(x \sin x)\)

⇒ \(y^{\prime}=\sin x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x) \Rightarrow y^{\prime}=\sin x+x \cos x\)

Substituting the value of y’ in the given differential equation, we get:

L.H.S. = x y’ = x(sin x + x cos x) = x sin x + x² cos x

= \(y+x^2 \cdot \sqrt{1-\sin ^2 x}=y+x^2 \sqrt{1-\left(\frac{y}{x}\right)^2}\) [Using Eq.(1)]

= \(y+x \sqrt{x^2-y^2}=\) R.H.S.

Hence, the given function is the solution of the corresponding differential equation

Question 7. \(x y=\log y+C: y^{\prime}=\frac{y^2}{1-x y} \quad(x y \neq 1)\)
Solution:

y – cos y = x….(1)

Differentiating both sides of the equation with respect to x, we get: \(\frac{d y}{d x}-\frac{d}{d x}(\cos y)=\frac{d}{d x}(x) \Rightarrow y^{\prime}+\sin y \cdot y^{\prime}\)=1

⇒ \(y^{\prime}(1+\sin y)=1 \Rightarrow y^{\prime}=\frac{1}{1+\sin y}\)….(2)

Substituting the value of y’ in the given differential equation, we get:

L.H.S. = \((y \sin y+\cos y+x) y^{\prime}=(y \sin y+\cos y+y-\cos y) \times \frac{1}{1+\sin y}\) [Using Eq.(1) and (2)]

= \(y(1+\sin y) \cdot \frac{1}{1+\sin y}=y=\text { R.H.S. }\)

Hence, the given function is the solution of the corresponding differential equation

Question 9. \(x+y=\tan ^{-1} y: y^2 y^{\prime}+y^2+1=0\)
Solution:

x+y = \(\tan ^{-1} y\)

Differentiating both sides of this equation with respect to x, we get:

⇒ \(\frac{d}{d x}(x+y)=\frac{d}{d x}\left(\tan ^{-1} y\right) \Rightarrow 1+y^{\prime}\)

= \(\left[\frac{1}{1+y^2}\right] y^{\prime} \Rightarrow y^{\prime}\left[\frac{1}{1+y^2}-1\right]=1\)

⇒ \(y^{\prime}\left[\frac{1-\left(1+y^2\right)}{1+y^2}\right]=1 \Rightarrow y^{\prime}\left[\frac{-y^2}{1+y^2}\right]\)=1

⇒ \(y^{\prime}=\frac{-\left(1+y^2\right)}{y^2}\)

Substituting the value of \(y^{\prime}\) in the given differential equation, \(y^2 y^1+y^2+1=0\) we get:

L.H.S. = \(y^2y^{\prime}+y^2+1=y^2\left[\frac{-\left(1+y^2\right)}{y^2}\right]+y^2+1=-1-y^2+y^2+1=0\)

Hence, the given function is the solution of the corresponding differential equation.

Question 10. y = \(\sqrt{a^2-x^2}, x \in(-a, a) \quad: x+y \frac{d y}{d x}=0(y \neq 0)\)
Solution:

y = \(\sqrt{a^2-x^2}\)….(1)

Differentiating both sides of this equation with respect to x, we get

⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{a^2-x^2}\right)\)

⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{a^2-x^2}} \cdot \frac{d}{d x}\left(a^2-x^2\right)=\frac{1}{2 \sqrt{a^2-x^2}}(-2 x)=\frac{-x}{\sqrt{a^2-x^2}}\)…(2)

Substituting the value of \(\frac{dy}{dx}\) in the given differential equation, we get

L.H.S. = \(x+y \frac{d y}{d x}=x+\sqrt{a^2-x^2} \times \frac{-x}{\sqrt{a^2-x^2}}\) (Using eq(1) and (2))

Hence, the given function is the solution of the corresponding differential equation.

Question 11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

1. 0
2. 2
3. 3
4. 4

Solution: 4. 4

We know that the number of constants in the general solution of a differential equation of order n is equal to its order.

Therefore, the number of constants in the general equation of fourth order differential equation is four.

Hence, the correct answer is (4).

Question 12. The number of arbitrary constants in the particular solution of a differential equation of third order are:

1. 3
2. 2
3. 1
4. 0

Solution: 4. 0

In a particular solution of a differential equation, there are no arbitrary constants.

Hence, the correct answer is (4).

Differential Equation Exercise 9.3

For Each Of The Differential Equations, Find The General Solution

Question 1. \(\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}\)
Solution:

The given differential equation is

⇒ \(\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x} \Rightarrow \frac{d y}{d x}=\frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\)

= \(\tan ^2 \frac{x}{2} \quad \Rightarrow \frac{d y}{d x}=\left(\sec ^2 \frac{x}{2}-1\right)\)

Separating the variables, we get: \(\mathrm{dy}=\left(\sec ^2 \frac{\mathrm{x}}{2}-1\right) \mathrm{dx}\)

Now, integrating both sides of this equation, we get:

⇒ \(\int d y=\int\left(\sec ^2 \frac{x}{2}-1\right) d x=\int \sec ^2 \frac{x}{2} d x-\int d x \Rightarrow y=2 \tan \frac{x}{2}-x+C\)

This is the required general solution of the given differential equation.

Question 3. \(\frac{d y}{d x}+y=1 \quad(y \neq 1)\)
Solution:

The given differential equation is: \(\frac{d y}{d x}+y=1 \Rightarrow \frac{d y}{d x}=1-y\)

Separating the variables, we get: \(\frac{d y}{1-y}=d x\)

Now, integrating both sides of this equation, we get: \(\int \frac{d y}{1-y}=\int d x\)

⇒ \(-\log (1-y)=x+\log C \Rightarrow-\log C-\log (1-y)=x \Rightarrow \log C(1-y)=-x\)

⇒ \(C(1-y)=e^{-x} \Rightarrow 1-y=\frac{1}{C} e^{-x} \Rightarrow y=1-\frac{1}{C} e^{-x} \Rightarrow y=1+A e^{-x}\)

(where  A = \(-\frac{1}{C}\))

This is the required general solution of the given differential equation.

Question 4. sec²x tan y dx + sec²y tan x dy =0
Solution:

The given differential equation is: sec²x tan y dx + sec²y tan x dy = 0

⇒ sec²x tan y dx = -sec²y tan x dy

On separating the variables, we get:

⇒ \(\frac{\sec ^2 x}{\tan x} d x=-\frac{\sec ^2 y}{\tan y} d y\)

Integrating both sides of this equation, we get: \(\int \frac{\sec ^2 x}{\tan x} d x=-\int \frac{\sec ^2 y}{\tan y} d y\)

⇒ log |tan x| = -log |tan y| = log C (\(\int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+C\))

⇒ log |tan x| + log |tan y| = log C

⇒ tan x tan y = C

This is the required general solution of the given differential equation.

Question 5. (e^x + e^-x)dy – (e^x – e^-x)dx = 0
Solution:

The given differential equation is: (e^x + e^-x)dy – (e^x – e^-x)dx = 0 ⇒ (e^x+ e^-x)dy = (e^x – e^-x)dx

Separating the variables, we get: \(d y=\left[\frac{e^x-e^{-x}}{e^x+e^{-x}}\right] d x\)

Integrating both sides of this equation, we get \(\int d y=\int\left[\frac{e^x-e^{-x}}{e^x+e^{-x}}\right] d x+C\)

⇒ y = \(\int\left[\frac{e^x-e^{-x}}{e^x+e^{-x}}\right] d x+C\)

⇒ y = \(\log \left(e^x+e^{-x}\right)+C\)

This is the required general solution of the given differential equation.

Question 6. \(\frac{d y}{d x}=\left(1+x^2\right)\left(1+y^2\right)\)
Solution:

The given differential equation is: \(\frac{d y}{d x}=\left(1+x^2\right)\left(1+y^2\right) \Rightarrow \frac{d y}{1+y^2}=\left(1+x^2\right) d x\)

Integrating both sides of this equation, we get:

⇒ \(\int \frac{d y}{1+y^2}=\int\left(1+x^2\right) d x \Rightarrow \tan ^{-1} y=\int d x+\int x^2 d x \Rightarrow \tan ^{-1} y=x+\frac{x^3}{3}+C\)

This is the required general solution of the given differential equation.

Question 7. y log y dx – x dy = 0
Solution:

The given differential equation is : y log y dx – x dy = 0 ⇒ y log y dx = x dy

Separating the variables, we get: \(\frac{d y}{y \log y}=\frac{d x}{x}\)

Integrating both sides, we get: \(\int \frac{d y}{y \log y}=\int \frac{d x}{x}\)….(1)

Let log y=t ⇒ \(\frac{1}{y} d y=d t\)

Substituting this value in equation (1), we get:

⇒ \(\int \frac{d t}{t}=\int \frac{d x}{x} \Rightarrow \log t=\log x+\log C \Rightarrow \log (\log y)=\log C x\)

⇒ \(\log y=C x \Rightarrow y=e^c\)

This is the required general solution of the given differential equation.

Question 8. \(x^5 \frac{d y}{d x}=-y^5\)
Solution:

The given differential equation is: \(x^5 \frac{d y}{d x}=-y^5\)

Separating the variables, we get: \(\frac{d y}{y^5}=-\frac{d x}{x^5} \Rightarrow \frac{d x}{x^5}+\frac{d y}{y^5}=0\)

Integrating both sides, we get: \(\int \frac{d x}{x^5}+\int \frac{d y}{y^5}=k\) where k is any constant

⇒ \(\int x^{-5} d x+\int y^{-5} d y=k \Rightarrow \frac{x^{-4}}{-4}+\frac{y^{-4}}{-4}=k \Rightarrow x^{-4}+y^{-4}=-4 k\)

⇒ \(x^{-4}+y^{-4}=C \quad(C=-4 k)\)

This is the required general solution of the given differential equation.

Question 9. \(\frac{d y}{d x}=\sin ^{-1} x\)
Solution:

The given differential equation \(\frac{d y}{d x}=\sin ^{-1} x \Rightarrow d y=\sin ^{-1} x d x\)

Integrating both sides, we get Integrating both sides, we get : \(\int d y=\int \sin ^{-1} x d x \Rightarrow y=\int\left(\sin ^{-1} x\right) \cdot 1 d x\)

⇒y = \(\sin ^{-1} x \cdot \int(1) d x-\int\left[\left(\frac{d}{d x}\left(\sin ^{-1} x\right) \cdot \int(1) d x\right)\right] d x\)

⇒ \(y=\sin ^{-1} x \cdot x-\int\left(\frac{1}{\sqrt{1-x^2}} \cdot x\right) d x\)

⇒ y = \(x \sin ^{-1} x+\int \frac{-x}{\sqrt{1-x^2}} d x\)….(1)

Let \(1-x^2=t \Rightarrow-2 x d x=d t\)

Substituting these values in equation (1), we get

y = \(x \sin ^{-1} x+\int \frac{1}{2 \sqrt{t}} d t \Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \cdot \int(t)^{-\frac{1}{2}} d t\)

⇒ \(y=x \sin ^{-1} x+\frac{1}{2} \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C\)]

⇒ y = \(x \sin ^{-1} x+\sqrt{t}+C \Rightarrow y=x \sin ^{-1} x+\sqrt{1-x^2}+C\)

This is the required general solution of the given differential equation.

Question 10. e^x tan y dx + (1-e^x) sec²ydy = 0
Solution:

The given differential equation is: \(e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0 \Rightarrow\left(1-e^x\right) \sec ^2 y d y=-e^x \tan y d x\)

Separating the variables, we get \(\frac{\sec ^2 y}{\tan y} d y=\frac{-e^x}{1-e^x} d x\)

Integrating both sides, we get: \(\int \frac{\sec ^2 y}{\tan y} d y=\int \frac{-e^x}{1-e^x} d x\)

⇒ \(\log (\tan y)=\log \left(1-e^x\right)+\log C\) (because \(\int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+C\))

⇒ \(\log (\tan y)=\log \left[C\left(1-e^x\right)\right]\)

⇒ \(\tan y=C\left(1-e^x\right)\)

This is the required general solution of the given differential equation.

For each of the differential equations, find a particular solution satisfying the given condition

Question 11. \(\left(x^3+x^2+x+1\right) \frac{d y}{d x}=2 x^2+x ; y=1\) when x=0
Solution:

The given differential equation is : \(\left(x^3+x^2+x+1\right) \frac{d y}{d x}=2 x^2+x \Rightarrow \frac{d y}{d x}=\frac{2 x^2+x}{\left(x^3+x^2+x+1\right)}\)

⇒ d y = \(\frac{2 x^2+x}{(x+1)\left(x^2+1\right)} d x\)

Integrating both sides, we get: \(\int d y=\int \frac{2 x^2+x}{(x+1)\left(x^2+1\right)} d x\)…..(1)

Let \(\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2+1}\)……(2)

⇒ \(\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A x^2+A+(B x+C)(x+1)}{(x+1)\left(x^2+1\right)}\)

⇒ \(2 x^2+x=A x^2+A+B x^2+B x+C x+C\)

⇒ \(2 x^2+x=(A+B) x^2+(B+C) x+(A+C)\)

Comparing the coefficients of x² and x, we get:

A + B = 2, B + C = 1,A + C = 0

Solving these equations, we get: A = 1/2, B = 3/2 and C = -1/2

Substituting the values of A, B, and C in equation (2), we get:

⇒ \(\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{1}{2} \cdot \frac{1}{(x+1)}+\frac{1}{2} \frac{(3 x-1)}{\left(x^2+1\right)}\)

Therefore, equation (1) becomes:

⇒ \(\int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^2+1} d x\)

⇒ \(y=\frac{1}{2} \log (x+1)+\frac{3}{2} \int \frac{x}{x^2+1} d x-\frac{1}{2} \int \frac{1}{x^2+1} d x\)

⇒ \(y=\frac{1}{2} \log (x+1)+\frac{3}{4} \cdot \int \frac{2 x}{x^2+1} d x-\frac{1}{2} \tan ^{-1} x+C\)

⇒ \(y=\frac{1}{2} \log (x+1)+\frac{3}{4} \log \left(x^2+1\right)-\frac{1}{2} \tan ^{-1} x+C \)

⇒ \(y=\frac{1}{4}\left[2 \log (x+1)+3 \log \left(x^2+1\right)\right]-\frac{1}{2} \tan ^{-1} x+C\)

⇒ \(y=\frac{1}{4}\left[\log (x+1)^2\left(x^2+1\right)^3\right]-\frac{1}{2} \tan ^{-1} x+C\)…(3)

Now, y=1 when x=0

⇒ \(1=\frac{1}{4} \log (1)-\frac{1}{2} \tan ^{-1} 0+C \Rightarrow 1=\frac{1}{4} \times 0-\frac{1}{2} \times 0+C \Rightarrow C=1\)

Substituting C=1 in equation (3), we get:

y = \(\frac{1}{4}\left[\log (x+1)^2\left(x^2+1\right)^3\right]-\frac{1}{2} \tan ^{-1} x+1\)

which is the required particular solution.

Question 12. \(x\left(x^2-1\right) \frac{d y}{d x}=1 ; y=0\) when x=2
Solution:

⇒ \(x\left(x^2-1\right) \frac{d y}{d x}=1\)

On separating the variables, we get : \(d y=\frac{d x}{x\left(x^2-1\right)} \Rightarrow \int d y=\int \frac{d x}{x^3\left(1-\frac{1}{x^2}\right)}\)

Put \(1-\frac{1}{x^2}=t \Rightarrow \frac{2}{x^3} d x=d t\),

⇒\(\int \mathrm{dy}=\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}} \Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{t})+\log \mathrm{C}\)

⇒ \(\mathrm{y}=\frac{1}{2} \log \left|\frac{\mathrm{x}^2-1}{\mathrm{x}^2}\right|+\log \mathrm{C}\)…(1)

Now, y = 0 when x = 2

0 = \(\frac{1}{2} \log \left(\frac{3}{4}\right)+\log \mathrm{C}\) (From eq.(1))

(\(\log \mathrm{C}=-\frac{1}{2} \log \left(\frac{3}{4}\right)\))

log \(\mathrm{C}=\frac{1}{2} \log \left(\frac{4}{3}\right)\)

⇒ \(\mathrm{y}=\frac{1}{2} \log \left(\frac{\mathrm{x}^2-1}{\mathrm{x}^2}\right)+\frac{1}{2} \log \left(\frac{4}{3}\right)\)

⇒ \(\mathrm{y}=\frac{1}{2} \log \left[\frac{4\left(\mathrm{x}^2-1\right)}{3 \mathrm{x}^2}\right]\)

which is the required particular solution.

Question 13. \(\cos \left(\frac{d y}{d x}\right)=a(a \in R) ; y=1\) when x=0
Solution:

cos \(\left(\frac{d y}{d x}\right)=a \Rightarrow \frac{d y}{d x}=\cos ^{-1} a\)

On separating the variables, we get, dy = cos^-1 a dx

Integrating both sides, we get: ∫dy = cos^-1 a ∫dx

⇒ y = cos^-1 a x + C ⇒ y = x cos^-1 a + C….(1)

Now, y = 1 when x = 0 ⇒ 1 = 0 · cos^-1 a + C ⇒ C = 1

Substituting C= 1 in equation (1), we get:

y = \(x \cos ^{-1} a+1 \Rightarrow \frac{y-1}{x}=\cos ^{-1} a \Rightarrow \cos \left(\frac{y-1}{x}\right)=a\)

which is the required particular solution.

Question 14. \(\frac{dy}{dx}\)= y tan x ; y = 1 when x = 0
Solution:

⇒ \(\frac{dy}{dx}\)= y tan x

Separating the variables, we get: \(\frac{dy}{dx}\) = tan x dx

Integrating both sides, we get: ∫\(\frac{dy}{dx}\) = ∫tan x dx

⇒ log y = log |sec x| + log C

⇒ log y = log(C sec x) ⇒ y C sec x……(1)

Now, y = 1 when x = 0 ⇒ 1= C x sec0 ⇒ 1 = C x 1 ⇒ C = 1

Substituting C = 1 in equation (1), we get y = sec x which is the required particular solution.

Question 15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y’ = e^x sin x
Solution:

The differential equation of the curve is: y’ = e^x sin x

⇒ \(\frac{dy}{dx}\) = e^x sin x

Separating the variables, we get: dy = e^x sin x dx

Integrating both sides, we get: J∫dy =∫e^x sin x dx….(1)

Let I = \(\int{e^x} \sin x d x\)

⇒ \(I=\sin x \int e^x d x-\int\left(\frac{d}{d x}(\sin x) \cdot \int e^x d x\right) d x \Rightarrow I=\sin x \cdot e^x-\int \cos x \cdot e^x d x\)

⇒ \(I=\sin x \cdot e^x-\left[\cos x \cdot \int e^x d x-\int\left(\frac{d}{d x}(\cos x) \cdot \int e^x d x\right) d x\right]\)

⇒ \(I=\sin x e^x-\left[\cos x \cdot e^x-\int(-\sin x) \cdot e^x d x\right] \Rightarrow I=e^x \sin x-e^x \cos x-I\)

⇒ \(2 I=e^x(\sin x-\cos x) \Rightarrow I=\frac{e^x(\sin x-\cos x)}{2}\)

Substituting this value in equation (1), we get:

y = \(\frac{e^x(\sin x-\cos x)}{2}+C\)…(2)

Now, the curve passes through point (0,0).

∴ \(0=\frac{\mathrm{e}^0(\sin 0-\cos 0)}{2}+\mathrm{C} \Rightarrow 0=\frac{1(0-1)}{2}+\mathrm{C} \Rightarrow \mathrm{C}=\frac{1}{2}\)

Substituting C = \(\frac{1}{2}\) in equation (2), we get:

y = \(=\frac{\mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}-\cos \mathrm{x})}{2}+\frac{1}{2}\)

⇒ \(2 \mathrm{y}=\mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}-\cos \mathrm{x})+1 \Rightarrow 2 \mathrm{y}-1=\mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}-\cos \mathrm{x})\)

Hence, the required equation of the curve is \(2 y-1=e^x(\sin x-\cos x)\)

Question 16. For the differential equation xy\(\frac{dy}{dx}\) = (x + 2)(y + 2). find the solution curve passing through the point (1,-1).
Solution:

The differential equation of the given curve is: xy \(\frac{dy}{dx}\)= (x + 2)(y + 2)

Separating the variables, we get: \(\left(\frac{y}{y+2}\right) d y=\left(\frac{x+2}{x}\right) d x \Rightarrow\left(1-\frac{2}{y+2}\right) d y=\left(1+\frac{2}{x}\right) d x\)

Integrating both sides, we get:

⇒ \(\int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x \Rightarrow \int d y-2 \int \frac{1}{y+2} d y=\int d x+2 \int \frac{1}{x} d x\)

⇒ \( y-2 \log (y+2)=x+2 \log x+C \Rightarrow y-x-C=\log x^2+\log (y+2)^2\)

⇒ \(y-x-C=\log \left[x^2(y+2)^2\right]\)…(1)

Now, the curve passes through point (1,-1).

⇒ \(-1-1-\mathrm{C}=\log \left[(1)^2(-1+2)^2\right] \Rightarrow-2-\mathrm{C}=\log 1=0 \Rightarrow \mathrm{C}=-2\)

Substituting C=-2 in equation (1), we get : \(y-x+2=\log \left[x^2(y+2)^2\right]\)

This is the required solution of the given curve.

Question 17. Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.
Solution:

Let x and y be the x-coordinate and y-coordinate of the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the relation, \(\frac{dy}{dx}\)

According to the given information, we get:

The product of the slope of a tangent with y-coordinate = x-coordinate

y · \(\frac{dy}{dx}\) = x

Separating the variables, we get; y dy = x dx

Integrating both sides, we get: \(\int y d y=\int x d x \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+C \Rightarrow y^2-x^2=2 C\)….(1)

Now, the curve passes through the point (0, -2).

∴ (-2)² – 0² = 2C ⇒ 2C = 4

Substituting 2C = 4 in equation (1), we get: y² – x² = 4

This is the required equation of the curve.

Question 18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). Find the equation of the curve given that it passes through (-2, 1).
Solution:

It is given that (x, y) is the point of contact of the curve and its tangent.

The slope (m₁) of the line segment joining (x, y) and (-4, -3) is \(\frac{y-(-3)}{x-(-4)}=\frac{y+3}{x+4}\)

We know that the slope of the tangent to the curve is given by the relation, \(\frac{dy}{dx}\)

∴ Slope (m₂) of the tangent = \(\frac{dy}{dx}\)

According to the given information : \(\mathrm{m}_2=2 \mathrm{~m}_1 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2(\mathrm{y}+3)}{\mathrm{x}+4}\)

Separating the variables, we get: \(\frac{d y}{y+3}=\frac{2 d x}{x+4}\)

Integrating both sides, we get: \(\int \frac{d y}{y+3}=2 \int \frac{d x}{x+4} \Rightarrow \log (y+3)=2 \log (x+4)+\log C\)

⇒ log(y + 3) = log C (x + 4)² ⇒ y + 3 = C(x + 4)²….(1)

This is the general equation of the curve.

It is given that it passes through the point (-2, 1).

⇒ 1 + 3 = C(-2 + 4)² ⇒ 4 = 4C ⇒ C = 1

Substituting C = 1 in equation (1), we get; y + 3 = (x + 4)²

This is the required equation of the curve.

Question 19. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.
Solution:

Let the rate of change of the volume of the balloon be k (where k is a constant).

⇒ \(\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{k} \Rightarrow \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{4}{3} \pi \mathrm{r}^3\right)=\mathrm{k}\)

(Volume of sphere = \(\frac{4}{3} \pi \mathrm{r}^3\))

⇒ \(\frac{4}{3} \pi \cdot 3 \mathrm{r}^2 \cdot \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k} \Rightarrow 4 \pi \mathrm{r}^2 \mathrm{dr}=\mathrm{kdt}\)

Integrating both sides, we get: \(4 \pi \int r^2 d r=k \int d t \Rightarrow 4 \pi \cdot \frac{r^3}{3}=k t+C \Rightarrow 4 \pi r^3=3(k t+C)\)….(1)

Now, at t=0, r=3

⇒ \(4 \pi \times 3^3=3(\mathrm{k} \times 0+\mathrm{C}) \Rightarrow 108 \pi=3 \mathrm{C} \Rightarrow \mathrm{C}=36 \pi\)

At t=3, r=6

⇒ \(4 \pi \times 6^3=3(\mathrm{k} \times 3+\mathrm{C}) \Rightarrow 864 \pi=3(3 \mathrm{k}+36 \pi) \Rightarrow 3 \mathrm{k}=288 \pi-36 \pi=252 \pi\)

⇒ k=84 \(\pi\)

Substituting the values of k and C in equation (1), we get:

⇒ \(4 \pi r^3=3[84 \pi t+36 \pi] \Rightarrow 4 \pi r^3=4 \pi[63 t+27] \Rightarrow r^3=63 t+27 \Rightarrow r=(63 t+27)^{\frac{1}{3}}\)

Thus, the radius of the balloon after t seconds is \((63 t+27)^{\frac{1}{3}}\).

Question 20. In a bank, the principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (log_e2 = 0.6931).
Solution:

Let p, t, and r represent the principal, time, and rate of interest respectively.

It is given that the principal increases continuously at the rate of r% per year.

⇒ \(\frac{d p}{d t}=\left(\frac{r}{100}\right) p \Rightarrow \frac{d p}{p}=\left(\frac{r}{100}\right) d t\)

Integrating both sides, we get:

⇒ \(\int \frac{d p}{p}=\frac{r}{100} \int d t \Rightarrow \log p=\frac{r t}{100}+k \Rightarrow p=e^{\frac{rt}{100}+k}\)….(1)

It is given that when \(\mathrm{t}=0, \mathrm{p}=100. \Rightarrow 100=\mathrm{e}^{\mathrm{k}}\)…..(2)

Now, if t=10, then p=2 x 100=200.

Therefore, equation (1) becomes :

200 = \(e^{\frac{\mathrm{t}}{10}+\mathrm{k}} \Rightarrow 200=\mathrm{e}^{\frac{\mathrm{r}}{10}} \cdot \mathrm{e}^{\mathrm{k}} \Rightarrow 200=\mathrm{e}^{\frac{\mathrm{t}}{11}} \cdot 100\)(From (2)

⇒ \(\mathrm{e}^{\frac{r}{10}}=2 \Rightarrow \frac{r}{10}=\log _4 2 \Rightarrow \frac{r}{10}=0.6931 \Rightarrow r=6.931\)

Hence, the value of r is 6.93%.

Question 21. In a bank, the principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it be worth after 10 years (e^0.5 = 1.648)?
Solution:

Let p and t be the principal and time respectively.

It is given that the principal increases continuously at the rate of 5% per year.

⇒ \(\frac{d p}{d t}=\left(\frac{5}{100}\right) p \Rightarrow \frac{d p}{d t}=\frac{p}{20}\)

Separating the variables, we get: \(\frac{d p}{d t}=\frac{dt}{20}\)

Integrating both sides, we get: \(\int \frac{d p}{p}=\frac{1}{20} \int d t \Rightarrow \log p=\frac{t}{20}+C \Rightarrow p=e^{\frac{t}{20}+C}\)…….(1)

Now, when t = 0, p = 1000.

⇒ 1000 = e^c

At t = 10, equation (1) becomes p=e^{1/2+ C} ⇒ p = e^0.5 x e^C ⇒ p = 1.648 x 1000 (from (2))

⇒ p = 1648

Hence, after 10 years the amount will be worth Rs 1648.

Question 22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00.000, if the rate of growth of bacteria is proportional to the number present?
Solution:

Let y be the number of bacteria at any instant t.

It is given that the rate of growth of the bacteria is proportional to the number present.

∴ \(\frac{d y}{d t} \propto y \Rightarrow \frac{d y}{d t}=k y\) (where k is constant)

Separating the variables, we get: \(\frac{d y}{d t}\) = k dt

Integrating both sides, we get: \(\frac{d y}{d t}\) = k dt ⇒ log y = kt + C……(1)

Let y₀ be the number of bacteria at t = 0. ⇒ log y₀ = C

Substituting the value of C in equation (1), we get:

log y = \(k t+\log y_0 \Rightarrow \log y-\log y_0=k t\)

⇒ \(\log \left(\frac{y}{y_0}\right)=k t\)…..(2)

Also, it is given that the number of bacteria increases by 10 % in 2 hours.

⇒ \(y=\frac{110}{100} y_0 \Rightarrow \frac{y}{y_0}=\frac{11}{10}\)….(3)

Substituting this value from equation (3) in the equation

⇒ \(\mathrm{k} \cdot 2=\log \left(\frac{11}{10}\right) \Rightarrow \mathrm{k}=\frac{1}{2} \log \left(\frac{11}{10}\right)\)

Therefore, equation (2) becomes:

⇒ \(\frac{1}{2} \log \left(\frac{11}{10}\right) \cdot t=\log \left(\frac{y}{y_0}\right) \Rightarrow t=\frac{2 \log \left(\frac{y}{y_0}\right)}{\log \left(\frac{11}{10}\right)}\)…..(4)

Now, let the time when the number of bacteria increases from 100000 to 200000 be t.

⇒ \(\mathrm{y}=2 \mathrm{y}_0\) at \(\mathrm{t}=\mathrm{t}_1\)

From equation (4), we get \(t_1=\frac{2 \log \left(\frac{y}{y_0}\right)}{\log \left(\frac{11}{10}\right)}=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}\).

Hence, in \(\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}\) hours the number of bacteria increases from 100000 to 200000.

Question 23. The general solution of the differential equation \(\frac{dy}{dt}\) = e^x+y is

1. \(e^x+e^{-y}=C\)
2. \(e^x+e^y=C\)
3. \(\mathrm{e}^{-x}+\mathrm{e}^y=\mathrm{C}\)
4. \(\mathrm{e}^{-x}+\mathrm{e}^{-7}=\mathrm{C}\)

Solution: 1. \(e^x+e^{-y}=C\)

Given, \(\frac{d y}{d x}=e^{x+y}=e^x \cdot e^y\)

Separating the variables, we get : \(\frac{d y}{e^y}=e^x d x \Rightarrow e^{-y} d y=e^x d x\)

Integrating both sides, we get:

⇒ \(\int \mathrm{e}^{-y} \mathrm{dy}=\int \mathrm{e}^{\mathrm{x}} \mathrm{dx} \Rightarrow-\mathrm{e}^{-\mathrm{y}}=\mathrm{e}^{\mathrm{x}}+\mathrm{k} \Rightarrow \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{y}}=-\mathrm{k} \Rightarrow \mathrm{e}^x+\mathrm{e}^{-\mathrm{y}}=\mathrm{C} \quad(\mathrm{C}=-\mathrm{k})\)

Hence, the correct answer is 1.

Differential Equations Exercise 9.4

Show That The Given Differential Equation Is Homogeneous And Solve Each Of Them.

Question 1. (x² + xy)dy = (x² + y²)dx
Solution:

The given differential equation i.e., (x² + xy)dy = (x² + y²)dx can be written as:

⇒ \(\frac{d y}{d x}=\frac{x^2+y^2}{x^2+x y}\)….(1)

Let F(x, y) = \(\frac{x^2+y^2}{x^2+x y}\)

Now, \(F(\lambda x, \lambda y)=\frac{(\lambda x)^2+\left(\lambda y^2\right)}{(\lambda x)^2+(\lambda x)(\lambda y)}=\frac{\lambda^2\left(x^2+y^2\right)}{\lambda^2\left(x^2+x y\right)}=\lambda^0 \cdot F(x, y)\)

This shows that equation (1) is a homogeneous equation.

To solve it, we make the substitution as y = vx

Differentiating both sides with respect to x. we get: \(\frac{dy}{dx}\) = v + x \(\frac{dv}{dx}\)

Substituting the values of y and \(\frac{dv}{dx}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{x^2+(v x)^2}{x^2+x(v x)} \Rightarrow v+x \frac{d v}{d x}=\frac{1+v^2}{1+v}\)

⇒ \(x \frac{d v}{d x}=\frac{1+v^2}{1+v}-v=\frac{\left(1+v^2\right)-v(1+v)}{1+v} \Rightarrow x \frac{d v}{d x}=\frac{1-v}{1+v}\)

⇒ \(\left(\frac{1+v}{1-v}\right) d v=\frac{d x}{x} \Rightarrow\left(\frac{2-1+v}{1-v}\right) d v=\frac{d x}{x} \Rightarrow\left(\frac{2}{1-v}-1\right) d v=\frac{d x}{x}\)

Integrating both sides, we get:-2 log (1-v)-v = log x-log k

⇒ v = -2 log (1-v)-log x+log k

⇒ v = \(\log \left[\frac{k}{x(1-v)^2}\right] \Rightarrow \frac{y}{x}=\log \left[\frac{k}{x\left(1-\frac{y}{x}\right)^2}\right] \Rightarrow \frac{y}{x}=\log \left[\frac{k x}{(x-y)^2}\right]\)

⇒ \(\frac{k x}{(x-y)^2}=e^{\frac{y}{x}} \Rightarrow(x-y)^2=k x e^{-\frac{x}{x}}\)

This is the required solution of the given differential equation.

Question 2. \(y^{\prime}=\frac{x+y}{x}\)
Solution:

The given differential equation is: \(y^{\prime}=\frac{x+y}{x} \Rightarrow \frac{d y}{d x}=\frac{x+y}{x}\)….(1)

Let F(x, y)= \(\frac{x+y}{x}\)

Now, \(\mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y})=\frac{\lambda \mathrm{x}+\lambda \mathrm{y}}{\lambda \mathrm{x}}=\frac{\lambda(\mathrm{x}+\mathrm{y})}{\lambda \mathrm{x}}=\lambda^0 \mathrm{~F}(\mathrm{x}, \mathrm{y})\)

Thus, the given equation is a homogeneous equation.

To solve it, we make the substitution as y = vx

Substituting the values of y and \(\frac{dy}{dx}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{x+v x}{x} \Rightarrow v+x \frac{d v}{d x}=1+v \Rightarrow x \frac{d v}{d x}=1 \Rightarrow \int d v=\int \frac{d x}{x}\)

Integrating both sides, we get:

v = \(\log x+C \Rightarrow \frac{y}{x}=\log x+C \Rightarrow y=x \log x+C x\)

This is the required solution of the given differential equation.

Question 3. (x – y) dy – (x + y)dx = 0
Solution:

The given differential equation is : (x – y) dy – (x + y)dx = 0

⇒ \(\frac{d y}{d x}=\frac{x+y}{x-y}\)…..(1)

Let F(x, y)= \(\frac{x+y}{x-y}\)

∴ \(\mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y})=\frac{\lambda \mathrm{x}+\lambda \mathrm{y}}{\lambda \mathrm{x}-\lambda \mathrm{y}}=\frac{\lambda(\mathrm{x}+\mathrm{y})}{\lambda(\mathrm{x}-\mathrm{y})}=\lambda^0 \cdot \mathrm{F}(\mathrm{x}, \mathrm{y})\)

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as : \(\mathrm{y}=\mathrm{vx}\)

⇒ \(\frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d y}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{x+v x}{x-v x}=\frac{1+v}{1-v} \Rightarrow x \frac{d v}{d x}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}\)

x \(\frac{d v}{d x}=\frac{1+v^2}{1-v} \Rightarrow \int \frac{1-v}{\left(1+v^2\right)} d v=\int \frac{d x}{x} \Rightarrow \int\left(\frac{1}{1+v^2}-\frac{v}{1+v^2}\right) d v\)

= \(\int \frac{d x}{x}\)

⇒ tan \(^{-1} v-\frac{1}{2} \log \left(1+v^2\right)=\log x+C \Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2} \log \left[1+\left(\frac{y}{x}\right)^2\right]=\log x+C\)

⇒ \(\tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2} \log \left(\frac{x^2+y^2}{x^2}\right)=\log x+C\)

⇒ \(\tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\left[\log \left(x^2+y^2\right)-\log x^2\right]=\log x+C\)

⇒ \(\tan ^{-1}\left(\frac{y}{x}\right)=\frac{1}{2} \log \left(x^2+y^2\right)+C\)

This is the required solution of the given differential equation.

Question 4. (x² – y²)dx + 2xy dy = 0
Solution:

The given differential equation is

⇒ \(\left(x^2-y^2\right) d x+2 x y d y=0 \Rightarrow \frac{d y}{d x}=\frac{-\left(x^2-y^2\right)}{2 x y}\)…..(1)

Let F(x, y) = \(\frac{-\left(x^2-y^2\right)}{2 x y}\)

∴ \(F(\lambda x, \lambda y)=-\left[\frac{(\lambda x)^2-(\lambda y)^2}{2(\lambda x)(\lambda y)}\right]=\frac{-\lambda^2\left(x^2-y^2\right)}{\lambda^2(2 x y)}=\lambda^0 \cdot F(x, y)\)

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = \(v x \Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=-\left[\frac{x^2-(v x)^2}{2 x \cdot(v x)}\right] \Rightarrow v+x \frac{d v}{d x}=\frac{v^2-1}{2 v}\)

⇒ \(x \frac{d v}{d x}=\frac{v^2-1}{2 v}-v=\frac{v^2-1-2 v^2}{2 v} \Rightarrow x \frac{d v}{d x}=-\frac{\left(1+v^2\right)}{2 v}\)

⇒ \(\frac{2 v}{1+v^2} d v=-\frac{d x}{x}\)

⇒ \(\int \frac{2 v}{1+v^2} d v=-\int \frac{d x}{x}\)

log \(\left(1+v^2\right)=-\log x+\log C=\log \frac{C}{x} \Rightarrow 1+v^2=\frac{C}{x} \Rightarrow\left[1+\frac{y^2}{x^2}\right]\)

= \(\frac{C}{x} \Rightarrow x^2+y^2=C x\)

This is the required solution of the given differential equation.

Question 5. \(x^2 \frac{d y}{d x}=x^2-2 y^2+x y\)
Solution:

The given differential equation is : \(x^2 \frac{d y}{d x}=x^2-2 y^2+x y\)

⇒ \(\frac{d y}{d x}=\frac{x^2-2 y^2+x y}{x^2}\)….(1)

Let F(x, y) = \(\frac{x^2-2 y^2+x y}{x^2}\)

∴ \(F(\lambda x, \lambda y)=\frac{(\lambda x)^2-2(\lambda y)^2+(\lambda x)(\lambda y)}{(\lambda x)^2}=\frac{\lambda^2\left(x^2-2 y^2+x y\right)}{\lambda^2\left(x^2\right)}=\lambda^0 \cdot F(x, y)\)

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = \(v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{x^2-2(v x)^2+x \cdot(v x)}{x^2}\)

⇒ \(v+x \frac{d v}{d x}=1-2 v^2+v \Rightarrow x \frac{d v}{d x}=1-2 v^2\)

⇒ \(\frac{d v}{1-2 v^2}=\frac{d x}{x} \Rightarrow \frac{1}{2} \int \frac{d v}{\frac{1}{2}-v^2}=\int \frac{d x}{x}\)

⇒ \(\frac{1}{2} \cdot \int\left[\frac{\mathrm{dv}}{\left(\frac{1}{\sqrt{2}}\right)^2-v^2}\right]=\int \frac{\mathrm{dx}}{\mathrm{x}}\)

⇒ \(\frac{1}{2} \cdot \frac{1}{2 \times \frac{1}{\sqrt{2}}} \log \left|\frac{\frac{1}{\sqrt{2}}+\mathrm{v}}{\frac{1}{\sqrt{2}}-\mathrm{v}}\right|=\log |\mathrm{x}|+\mathrm{C}\)

(because \(\int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\))

⇒ \(\frac{1}{2 \sqrt{2}} \log \left|\frac{\frac{1}{\sqrt{2}}+\frac{y}{x}}{\frac{1}{\sqrt{2}}-\frac{y}{x}}\right|=\log |x|+C \Rightarrow \frac{1}{2 \sqrt{2}} \log \left|\frac{x+\sqrt{2} y}{x-\sqrt{2} y}\right|=\log |x|+C \)

This is the required solution for the given differential equation.

Question 6. \(x d y-y d x=\sqrt{x^2+y^2} d x\)
Solution:

⇒ \(x d y-y d x=\sqrt{x^2+y^2} d x \Rightarrow x d y=\left[y+\sqrt{x^2+y^2}\right] d x\)

⇒ \(\frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x}\)….(1)

Let F(x, y) = \(\frac{y+\sqrt{x^2+y^2}}{x}\)

∴ \(F(\lambda x, \lambda y)=\frac{\lambda x+\sqrt{(\lambda x)^2+(\lambda y)^2}}{\lambda x}=\frac{\lambda\left(y+\sqrt{x^2+y^2}\right)}{\lambda x}=\lambda^0 \cdot F(x, y)\)

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as y = vx

⇒ \(\frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{v x+\sqrt{x^2+(v x)^2}}{x} \Rightarrow v+x \frac{d v}{d x}=v+\sqrt{1+v^2}\)

⇒ \(\frac{d v}{\sqrt{1+v^2}}=\frac{d x}{x} \Rightarrow \int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x}\)

⇒ \(\log \left|v+\sqrt{1+v^2}\right|=\log |x|+\log C \Rightarrow \log \left|\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right|=\log |C x|\)

⇒ \(\log \left|\frac{y+\sqrt{x^2+y^2}}{x}\right|=\log |C x| \Rightarrow y+\sqrt{x^2+y^2}=C x^2\)

This is the required solution of the given differential equation.

Question 7. \(\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y\)
Solution:

The given differential equation is:

⇒ \(\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y\)

⇒ \(\frac{d y}{d x}=\frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}\)….(1)

Let F(x, y) = \(\frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}\)

⇒ \(\mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y})=\frac{\left\{\lambda x \cos \left(\frac{\lambda y}{\lambda x}\right)+\lambda y \sin \left(\frac{\lambda y}{\lambda x}\right)\right\} \lambda y}{\left\{\lambda y \sin \left(\frac{\lambda y}{\lambda x}\right)-\lambda x \cos \left(\frac{\lambda y}{\lambda x}\right)\right\} \lambda x}\)

= \(\frac{\lambda^2\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\lambda^2\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}=\lambda^0 \cdot F(x, y)\)

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = \(v x \Rightarrow \frac{d y}{d x}=v+x \times \frac{d v}{d x}\)

Substituting the values of y and \(\frac{dy}{dx}\) in equation {1), we get:

v+x \(\frac{d v}{d x}=\frac{(x \cos v+v x \sin v) \cdot v x}{(v x \sin v-x \cos v) \cdot x} \Rightarrow v+x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v}\)

⇒ \(x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v}-v\)

⇒ \(x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v-v^2 \sin v+v \cos v}{v \sin v-\cos v}\)

⇒ \(x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v} \Rightarrow\left[\frac{v \sin v-\cos v}{v \cos v}\right] d v=\frac{2 d x}{x}\)

⇒ \(\left(\tan v-\frac{1}{v}\right) d v=\frac{2 d x}{x}\)

Integrating both sides, we get : \(\int\left(\tan v-\frac{1}{v}\right) d v=2 \int \frac{d x}{x}\)

log (sec v)-log v=2 log x+log C

⇒ \(\log \left(\frac{\sec v}{v}\right)=\log \left(C x^2\right) \Rightarrow\left(\frac{\sec v}{v}\right)=C^2 \Rightarrow \sec v=C^2 v\)

⇒ \(\sec \left(\frac{y}{x}\right)=C \cdot x^2 \cdot \frac{y}{x} \Rightarrow \sec \left(\frac{y}{x}\right)=C x y \Rightarrow \cos \left(\frac{y}{x}\right)=\frac{1}{C x y}=\frac{1}{C} \cdot \frac{1}{x y}\)

⇒ \(x y \cos \left(\frac{y}{x}\right)=k \quad\left(k=\frac{1}{C}\right)\)

This is the required solution of the given differential equation.

Question 8. \(x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0\)
Solution:

x \(\frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0 \Rightarrow x \frac{d y}{d x}=y-x \sin \left(\frac{y}{x}\right)\)

⇒ \(\frac{d y}{d x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}\)…..(1)

Let F(x, y) = \(\frac{y-x \sin \left(\frac{y}{x}\right)}{x}\)

⇒ \(\mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y})=\frac{\lambda \mathrm{y}-\lambda \mathrm{x} \sin \left(\frac{\lambda \mathrm{y}}{\lambda \mathrm{x}}\right)}{\lambda \mathrm{x}}=\frac{\lambda\left[\mathrm{y}-\mathrm{x} \sin \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right]}{\lambda \mathrm{x}}=\lambda^0 \cdot \mathrm{F}(\mathrm{x}, \mathrm{y})\)

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = \(v x \Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the values of x and \(\frac{dy}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{v x-x \sin v}{x} \Rightarrow v+x \frac{d v}{d x}=v-\sin v\)

⇒ \(-\frac{d v}{\sin v}=\frac{d x}{x} \Rightarrow \mathrm{cosec} v d v=-\frac{d x}{x} \Rightarrow \int \mathrm{cosec} v d v=-\int \frac{d x}{x}\)

log \(|\mathrm{cosec} v-\cot v|=-\log x+\log C=\log \frac{C}{x}\)

⇒ cosec \(\left(\frac{y}{x}\right)-\cot \left(\frac{y}{x}\right)=\frac{C}{x}\)

⇒ \(\frac{1}{\sin \left(\frac{y}{x}\right)}-\frac{\cos \left(\frac{y}{x}\right)}{\sin \left(\frac{y}{x}\right)}=\frac{C}{x}\)

⇒ x \(\left[1-\cos \left(\frac{y}{x}\right)\right]=C \sin \left(\frac{y}{x}\right)\)

This is the required solution of the given differential equation.

Question 9. \(y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0\)
Solution:

⇒ \(y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0 \Rightarrow y d x=\left[2 x-x \log \left(\frac{y}{x}\right)\right] d y\)

⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{2 \mathrm{x}-\mathrm{x} \log \left(\frac{\mathrm{y}}{\mathrm{x}}\right)}\)….(1)

Let F(x, y) = \(\frac{y}{2 x-x \log \left(\frac{y}{x}\right)}\)

∴ \(F(\lambda x, \lambda y)=\frac{\lambda y}{2(\lambda x)-(\lambda x) \log \left(\frac{\lambda y}{\lambda x}\right)}\)

= \(\frac{\lambda y}{\lambda\left[2 x-x \log \left(\frac{y}{x}\right)\right]}=\lambda^0 F \mathrm{~F}(x, y)\)

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: y = \(v x \Rightarrow \frac{d y}{d x}=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{v x}{2 x-x \log v} \Rightarrow v+x \frac{d v}{d x}=\frac{v}{2-\log v} \Rightarrow x \frac{d v}{d x}=\frac{v}{2-\log v}-v\)

⇒ \(x \frac{d v}{d x}=\frac{v-2 v+v \log v}{2-\log v} \Rightarrow x \frac{d v}{d x}=\frac{v \log v-v}{2-\log v} \Rightarrow \frac{2-\log v}{v(\log v-1)} d v=\frac{d x}{x}\)

⇒ \({\left[\frac{1+(1-\log v)}{v(\log v-1)}\right] d v=\frac{d x}{x} \Rightarrow\left[\frac{1}{v(\log v-1)}-\frac{1}{v}\right] d v=\frac{d x}{x} }\)

Integrating both sides, we get: \(\int \frac{1}{v(\log v-1)} d v-\int \frac{1}{v} d v=\int \frac{1}{x} d x\)

⇒ \(\int \frac{d v}{v(\log v-1)}-\log v=\log x+\log C\)…..(2)

Let log v-1=t

⇒ \(\frac{\mathrm{d}}{\mathrm{dv}}(\log \mathrm{v}-1)=\frac{\mathrm{dt}}{\mathrm{dv}} \Rightarrow \frac{1}{\mathrm{v}}=\frac{\mathrm{dt}}{\mathrm{dv}} \Rightarrow \frac{\mathrm{dv}}{\mathrm{v}}=\mathrm{dt}\)

Therefore, equation (2) becomes:

⇒ \(\int \frac{d t}{t}-\log v=\log x+\log C\)

⇒ \(\log t-\log v=\log (C x)=\log (\log v-1)-\log v=\log (C x)\)

⇒ \(\log \left[\log \left(\frac{y}{x}\right)-1\right]-\log \left(\frac{y}{x}\right)=\log (C x)\)

⇒ \(\log \left[\frac{\log \left(\frac{y}{x}\right)-1}{\frac{y}{x}}\right]=\log (C x) \Rightarrow \frac{x}{y}\left[\log \left(\frac{y}{x}\right)-1\right]=C x \Rightarrow \log \left(\frac{y}{x}\right)-1=C y\)

This is the required solution of the given differential equation.

Question 10. \(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0\)
Solution:

⇒ \(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0 \Rightarrow\left(1+e^{\frac{x}{y}}\right) d x=-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y\)

⇒ \(\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{-\mathrm{e}^{\frac{x}{y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right)}{1+\mathrm{e}^{\frac{\mathrm{x}}{y}}}\)….(1)

Let F(x, y) = \(\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}\)

∴ \(F(\lambda x, \lambda y)=\frac{-e^{\frac{\lambda x}{\lambda y}}\left(1-\frac{\lambda x}{\lambda y}\right)}{1+e^{\frac{\lambda x}{\lambda y}}}\)

= \(\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}=\lambda^0 \cdot F(x, y)\)

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: \(\mathrm{x}=\mathrm{vy} \Rightarrow \frac{\mathrm{d}}{\mathrm{dy}}(\mathrm{x})\)

= \(\frac{\mathrm{d}}{\mathrm{dy}}(\mathrm{vy}) \Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\mathrm{v}+\mathrm{y} \frac{\mathrm{dv}}{\mathrm{dy}}\)

Substituting the values of x and \(\frac{\mathrm{dx}}{\mathrm{dy}}\) in equation (1), we get:

v+y \(\frac{d v}{d y}=\frac{-e^v(1-v)}{1+e^v} \Rightarrow y \frac{d v}{d y}=\frac{-e^v+v^v}{1+e^v}-v\)

⇒ \(y \frac{d v}{d y}=\frac{-e^v+v^v-v-v e^v}{1+e^v} \Rightarrow y \frac{d v}{d y}=-\left[\frac{v+e^v}{1+e^v}\right]\)

⇒ \(\left[\frac{1+e^v}{v+e^v}\right] d v=-\frac{d y}{y}\)

Integrating both sides, we get:

⇒ \(\log \left(v+e^v\right)=-\log y+\log C=\log \left(\frac{C}{y}\right) \Rightarrow \log \left(v+e^v\right)=\log \left(\frac{C}{y}\right)\)

⇒ \(\frac{x}{y}+e^{\frac{x}{y}}=\frac{C}{y} \Rightarrow x+y e^y=C\)

This is the required solution of the given differential equation.

For Each Of The Differential Equations, Find The Particular Solution Satisfying The Given Condition:

Question 11. (x+y) d y+(x-y) d x=0 ; y=1 when x=1
Solution:

(x+y) d y+(x-y) d x=0 \Rightarrow(x+y) d y=-(x-y) d x

⇒ \(\frac{d y}{d x}=\frac{-(x-y)}{x+y}\)………(1)

The given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = \(v x \Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

⇒ \(v+x \frac{d v}{d x}=\frac{-(x-v x)}{x+v x} \Rightarrow v+x \frac{d v}{d x}=\frac{v-1}{v+1}\)

⇒ \(x \frac{d v}{d x}=\frac{v-1}{v+1}-v=\frac{v-1-v(v+1)}{v+1} \Rightarrow x \frac{d v}{d x}=\frac{v-1-v^2-v}{v+1}=\frac{-\left(1+v^2\right)}{v+1}\)

⇒ \(\frac{(v+1)}{1+v^2} d v=-\frac{d x}{x} \Rightarrow \int \frac{v+1}{1+v^2} d v=-\int \frac{d x}{x}\)

⇒ \(\int\left[\frac{v}{1+v^2}+\frac{1}{1+v^2}\right] d v=-\int \frac{d x}{x}\)

⇒ \(\frac{1}{2} \log \left(1+v^2\right)+\tan ^{-1} v=-\log x+k \Rightarrow \log \left(1+v^2\right)+2 \tan ^{-1} v=-2 \log x+2 k\)

⇒ \(\log \left(x^2+y^2\right)+2 \tan ^{-1} \frac{y}{x}=2 k\)…..(2)

Now, y=1 at x=1

⇒ \(\log 2+2 \tan ^{-1} 1=2 k \Rightarrow \log 2+2 \times \frac{\pi}{4}=2 k \Rightarrow \frac{\pi}{2}+\log 2=2 k\)

Substituting the value of 2k in equation (2), we get:

⇒ \(\log \left(x^2+y^2\right)+2 \tan ^{-1}\left(\frac{y}{x}\right)=\frac{\pi}{2}+\log 2\)

This is the required solution of the given differential equation.

Question 12. \(x^2 d y+\left(x y+y^2\right) d x=0\) ; y=1 when x=1
Solution:

⇒ \(x^2 d y+\left(x y+y^2\right) d x=0 \Rightarrow x^2 d y=-\left(x y+y^2\right) d x \)

⇒ \(\frac{d y}{d x}=\frac{-\left(x y+y^2\right)}{x^2}\)….(1)

The given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx \(\frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{-\left[x \cdot v x+(v x)^2\right]}{x^2}=-v-v^2\)

⇒ \(x \frac{d v}{d x}=-v^2-2 v=-v(v+2) \Rightarrow \frac{d v}{v(v+2)}=-\frac{d x}{x}\)

⇒ \(\int \frac{d v}{v(v+2)}=-\int \frac{d x}{x} \Rightarrow \int \frac{1}{2}\left[\frac{(v+2)-v}{v(v+2)}\right] d v\)

= \(-\int \frac{d x}{x} \Rightarrow \int \frac{1}{2}\left[\frac{1}{v}-\frac{1}{v+2}\right] d v=-\int \frac{d x}{x}\)

⇒ \(\frac{1}{2}[\log v-\log (v+2)]=-\log x+\log C \Rightarrow \frac{1}{2} \log \left(\frac{v}{v+2}\right)=\log \frac{C}{x}\)

⇒ \(\frac{v}{v+2}=\left(\frac{C}{x}\right)^2 \Rightarrow \frac{\frac{y}{x}}{\frac{y}{x}+2}=\left(\frac{C}{x}\right)^2\)

⇒ \(\frac{y}{y+2 x}=\frac{C^2}{x^2} \Rightarrow \frac{x^2 y}{y+2 x}=C^2\)….(2)

Put, y=1 and x=1 in eq.(2), we get \(\frac{1}{1+2}=C^2 \Rightarrow C^2=\frac{1}{3}\)

Substituting \(C^2=\frac{1}{3}\) in equation (2), we get: \(\frac{x^2 y}{y+2 x}=\frac{1}{3} \Rightarrow y+2 x=3 x^2 y\)

This is the required solution of the given differential equation

Question 13. \(\left[x \sin ^2\left(\frac{y}{x}\right)-y\right] d x+x d y=0 ; y=\frac{\pi}{4}\) when x=1
Solution:

⇒ \(\left[x \sin ^2\left(\frac{y}{x}\right)-y\right] d x+x d y=0 \Rightarrow \frac{d y}{d x}\)

= \(\frac{-\left[x \sin ^2\left(\frac{y}{x}\right)-y\right]}{x}\)…….(1)

The given differential equation is a homogeneous equation.

To solve this differential equation, we make the substitution as: \(\mathrm{y}=\mathrm{vx} \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{y})=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{vx}) \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{-\left[x \sin ^2 v-v x\right]}{x} \Rightarrow v+x \frac{d v}{d x}=-\left[\sin ^2 v-v\right]=v-\sin ^2 v\)

⇒ \(x \frac{d v}{d x}=-\sin ^2 v \Rightarrow \frac{d v}{\sin ^2 v}=-\frac{d x}{x} \)

⇒ \(\mathrm{cosec}^2 v d v=-\frac{d x}{x} \Rightarrow \int \mathrm{cosec}^2 v d v=-\int \frac{d x}{x}\)

⇒ \(\cot \left(\frac{y}{x}\right)=\log |x|-\log C \Rightarrow \cot \left(\frac{y}{x}\right)=\log \left|\frac{x}{C}\right|\)…..(2)

Now, \(y=\frac{\pi}{4}\) at x=1 \(\Rightarrow \cot \left(\frac{\pi}{4}\right)=\log \left|\frac{1}{C}\right| \Rightarrow 1=-\log C \Rightarrow C=e^{-t}\)

Substituting \(\mathrm{C}=\mathrm{e}^{-1}\) in equation (2), we get : \(\cot \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\log |\mathrm{ex}|\)

This is the required solution of the given differential equation.

Question 14. \(\frac{d y}{d x}-\frac{y}{x}+\mathrm{cosec}\left(\frac{y}{x}\right)=0 ; y=0\) when x=1
Solution:

⇒ \(\frac{d y}{d x}=\frac{y}{x}-\mathrm{cosec}\left(\frac{y}{x}\right)\)…..(1)

The given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = \(v x \Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=v-\mathrm{cosec} v \Rightarrow-\frac{d v}{\mathrm{cosec} v}=\frac{d x}{x}\)

⇒ \(-\sin v d v=\frac{d x}{x} \Rightarrow-\int \sin v d v=\int \frac{d x}{x}\)

cos v = \(\log x+\log C=\log |C x| \Rightarrow \cos \left(\frac{y}{x}\right)=\log |C x|\)….(2)

This is the required solution of the given differential equation.

Now, y=0 at x=1

⇒ cos (0) = \(\log \mathrm{C} \Rightarrow 1=\log \mathrm{C} \Rightarrow \mathrm{C}=\mathrm{e}^{\prime}=\mathrm{e}\)

Substituting C=e in equation (2), we get: \(\cos \left(\frac{y}{x}\right)=\log |(e x)|\)

This is the required solution of the given differential equation.

Question 15. \(2 x y+y^2-2 x^2 \frac{d y}{d x}=0 ; y=2\) when x=1
Solution:

2xy + \(y^2-2 x^2 \frac{d y}{d x}=0 \Rightarrow 2 x^2 \frac{d y}{d x}=2 x y+y^2 \Rightarrow \frac{d y}{d x}=\frac{2 x y+y^2}{2 x^2}\)….(1)

The given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = \(v x \Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the value of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{2 x(v x)+(v x)^2}{2 x^2} \Rightarrow v+x \frac{d v}{d x}=\frac{2 v+v^2}{2} \Rightarrow v+x \frac{d v}{d x}=v+\frac{v^2}{2}\)

⇒ \(\frac{2}{v^2} d v=\frac{d x}{x} \Rightarrow \int \frac{2}{v^2} d v=\int \frac{d x}{x}\)

⇒ \(2 \cdot \frac{v^{-2 d}}{-2+1}=\log |x|+C \Rightarrow-\frac{2}{v}=\log |x|+C\)

⇒ \(-\frac{2}{y}=\log |x|+C \Rightarrow-\frac{2 x}{y}=\log |x|+C\)…..(2)

Now, y = 2 at x = l.

⇒ -l = log(1) + C ⇒ C = -1

Substituting C = -1 in equation (2), we get:

– \(\frac{2 x}{y}=\log |x|-1 \Rightarrow \frac{2 x}{y}=1-\log |x| \Rightarrow y=\frac{2 x}{1-\log |x|},(x \neq 0, x \neq e)\)

This is the required solution of the given differential equation.

Question 16. A homogeneous differential equation of the form \(\frac{dx}{dy}\) = h(\(\frac{x}{y}\)) can be solved by making the substitution

1. y = vx
2. v = yx
3. x = vy
4. x = v

Solution:

For solving the homogeneous equation of the form \(\frac{dx}{dy}\) = h(\(\frac{x}{y}\)), we need to make the substitution as x = vy.

Hence, the correct answer is 3.

Question 17. Which of the following is a homogeneous differential equation?

1. (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0
2. (xy)dx – (x³ + y³)dy = 0
3. (x³ + 2y²)dx + 2xydy = 0
4. y²dx + (x² – xy – y²)dy = 0

Solution:

Out of the given four options; option (4) is the only option in which all coefficients of dx and dy are of the same degree, therefore function F(x, y) is said to be the homogenous function of degree n, if

F(λx, λy) = λ” F(x, y) for any non-zero constant (X).

Consider the equation given in option D: y²dx + (x² – xy – y²)dy = 0

⇒ \(\frac{d y}{d x}=\frac{-y^2}{x^2-x y-y^2}=\frac{y^2}{y^2+x y-x^2}\)

Let F(x, y) = \(\frac{y^2}{y^2+x y-x^2} \Rightarrow F(\lambda x, \lambda y)=\frac{(\lambda y)^2}{(\lambda y)^2+(\lambda x)(\lambda y)-(\lambda x)^2}\)

= \(\frac{\lambda^2 y^2}{\lambda^2\left(y^2+x y-x^2\right)}=\lambda^0\left(\frac{y^2}{y^2+x y-x^2}\right)=\lambda^0 \cdot F(x, y)\)

Hence, the differential equation given in option 4 is a homogenous equation.

Differential Equation Exercise 9.5

For Each Of The Differential Equations, Find The General Solution

Question 1. \(\frac{d y}{d x}+2 y=\sin x\)
Solution:

The given differential equation is \(\frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y}=\sin \mathrm{x}\)

This is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\) (where P=2 and Q= sin x)

Now, I.F. = \(\mathrm{e}^{\int \mathrm{Pdtx}}=\mathrm{e}^{\int 2 \mathrm{dx}}=\mathrm{e}^{2 \mathrm{x}}\)

The solution of the given differential equation is given by the relation,

y(I.F) = \(\int(Q \times \text { I.F. }) d x+C \Rightarrow y e^{2 x}=\int \sin x \cdot e^{2 x} d x+C\)…..(1)

Let \(I=\int_I \sin x \cdot e^{2 x} d x\)

⇒ \(I=\sin x \cdot \int e^{2 x} d x-\int\left(\frac{d}{d x}(\sin x) \cdot \int e^{2 x} d x\right) d x\)

⇒ \(I=\sin x \cdot \frac{e^{2 x}}{2}-\int\left(\cos x \cdot \frac{e^{2 x}}{2}\right) d x\)

⇒ I = \(\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \int e^{2 x}-\int\left(\frac{d}{d x}(\cos x) \cdot \int e^{2 x} d x\right) d x\right]\)

⇒ \(I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \frac{e^{2 x}}{2}-\int\left[(-\sin x) \cdot \frac{e^{2 x}}{2}\right] d x\right]\)

⇒ I = \(\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} \int\left(\sin x \cdot e^{2 x}\right) d x\)

⇒ \(I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)-\frac{1}{4} I \Rightarrow \frac{5}{4} I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)\)

⇒ \(I=\frac{e^{2 x}}{5}(2 \sin x-\cos x)\)

Therefore, equation (1) becomes: \(\mathrm{ye}^{2 \mathrm{x}}=\frac{\mathrm{e}^{2 \mathrm{x}}}{5}(2 \sin \mathrm{x}-\cos \mathrm{x})+\mathrm{C} \Rightarrow \mathrm{y}=\frac{1}{5}(2 \sin \mathrm{x}-\cos \mathrm{x})+\mathrm{Ce}^{-2 \mathrm{x}}\)

This is the required general solution of the given differential equation.

Question 2. \(\frac{d y}{d x}+3 y=e^{-2 x}\)
Solution:

The given differential equation is \(\frac{d y}{d x}+3 y=e^{-2 x}\).

This is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\) (where P=3 and Q = \(e^{-2 x}\))

Now, I.F. = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{3 \mathrm{dx}}{\mathrm{s}}=\mathrm{e}^{3 \mathrm{x}}}\)

The solution of the given differential equation is given by the relation,

y(I.F.)= \(\int(Q \times \text { I.F. }) d x+C \Rightarrow y e^{3 x}=\int\left(e^{-2 x} \times e^{3 x}\right)+C \Rightarrow y e^{3 x}=\int e^x d x+C\)

⇒ \(y e^{3 x}=e^x+C \Rightarrow y=e^{-2 x}+C e^{-5 x}\)

This is the required general solution of the given differential equation.

Question 3. \(\frac{d y}{d x}+\frac{y}{x}=x^2\)
Solution:

The given differential equation is \(\frac{d y}{d x}+\frac{y}{x}=x^2\).

This is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\) (where \(P=\frac{1}{x}\) and \(Q=x^2\))

Now, I.F. = \(\mathrm{e}^{\int \text { Pdx }}=\mathrm{e}^{\int \frac{1}{x}-\mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)

The solution of the given differential equation is given by,

y(I.F.) = \(\int(Q \times I . F .) d x+C \Rightarrow y(x)=\int\left(x^2 \cdot x\right) d x+C\)

⇒ \(x y=\int x^3 d x+C \Rightarrow x y=\frac{x^4}{4}+C\)

This is the required general solution of the given differential equation.

Question 4. \(\frac{d y}{d x}+(\sec x) y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)\)
Solution:

The given differential equation is \(\frac{\mathrm{dy}}{\mathrm{dx}}+(\sec x) \mathrm{y}=\tan \mathrm{x}\).

This is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\) (where P=sec x and Q=tan x)

Now, I.F. = \(\mathrm{e}^{\int \mathrm{Prtx}}=\mathrm{e}^{\int \sec x d x}=\mathrm{e}^{\log (\sec x+\tan x)}=\sec \mathrm{x}+\tan \mathrm{x}\)

The general solution of the given differential equation is given by the relation,

y(I.F.) = \(\int(Q \times \text { I.F. }) d x+C\)

⇒ y(sec x+tan x) = \(\int \tan x(\sec x+\tan x) d x+C \Rightarrow y(\sec x+\tan x)=\int \sec x \tan x d x+\int \tan ^2 x d x+C\)

⇒ y(sec x+tan x) = \(\sec x+\int\left(\sec ^2 x-1\right) d x+C \Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C\)

Question 5. \(\cos ^2 x \frac{d y}{d x}+y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)\)
Solution:

It is given that \(\cos ^2 x \frac{d y}{d x}+y=\tan x \Rightarrow \frac{d y}{d x}+\sec ^2 x y=\sec ^2 x \tan x\)

This is a differential equation in the form of \(\frac{d y}{d x}+P y=Q\) (where, P= \(sec ^2 x\) and Q = \(\sec ^2 x \tan x\))

Now, I.F. = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \sec ^2 \mathrm{x} d \mathrm{x}}=\mathrm{e}^{\tan \mathrm{x}}\)

Thus, the solution of the given differential equation is given by the relation :

y(I.F.) = \(\int(Q \times I . F) d x+C\)

⇒ \(y \cdot e^{\mathrm{tan} x}=\int e^{\mathrm{tan} x} \sec ^2 x \tan x d x+C\)….(1)

Now, Let tan x = \(t \Rightarrow \sec ^2 x d x=d t\)

Thus, the equation (1) becomes

⇒ \(y \cdot e^{\tan x}=\int\left(e^t \cdot t\right) d t+C \Rightarrow y \cdot e^{\tan x}=t \cdot \int e^t d t-\int\left(\frac{d}{d t}(t) \cdot \int e^t d t\right) d t+C\)

⇒ \(y \cdot e^{\tan x}=t \cdot e^t-\int e^t d t+C \Rightarrow y e^{\tan x}=(t-1) e^t+C\)

⇒ \(y e^{\tan x}=(\tan x-1) e^{\tan x}+C \Rightarrow y=(\tan x-1)+C e^{-\tan x}\)

Therefore, the required general solution of the given differential equation \(y=(\tan x-1)+C e^{-\tan x}\)

Question 6. \(x \frac{d y}{d x}+2 y=x^2 \log x\)
Solution:

The given differential equation is : \(x \frac{d y}{d x}+2 y=x^2 \log x \Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x\)

This equation is in the form of a linear differential equation as: \(\frac{d y}{d x}\) +Py=Q (where P = \(\frac{2}{x}\) and Q = x log x)

Now, I.F. = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int_{\mathrm{x}}^2 \mathrm{dx}}=\mathrm{e}^{2 \log x}=\mathrm{e}^{\log \mathrm{x}^2}=\mathrm{x}^2\)

The general solution of the given differential equation is given by the relation,

y(I.F.) = \(\int(Q \times I . F .) d x+C \Rightarrow y \cdot x^2=\int\left(x \log x \cdot x^2\right) d x+C\)

⇒ \(x^2 y=\int\left(x^3 \log x\right) d x+C \Rightarrow x^2 y=\log x \cdot \int x^3 d x-\int\left[\frac{d}{d x}(\log x) \cdot \int x^3 d x\right] d x+C\)

⇒ \(x^2 y=\log x \cdot \frac{x^4}{4}-\int\left(\frac{1}{x} \frac{x^4}{4}\right) d x+C \Rightarrow x^2 y=\frac{x^4 \log x}{4}-\frac{1}{4} \int x^3 d x+C\)

⇒ \(x^2 y=\frac{x^4 \log x}{4}-\frac{1}{4} \cdot \frac{x^4}{4}+C \Rightarrow x^2 y=\frac{1}{16} x^4(4 \log x-1)+C\)

⇒ \(y=\frac{1}{16} x^2(4 \log x-1)+C x^{-2}\)

Question 7. \(x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x\)
Solution:

The given differential equation is: \(x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x \Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^2}\)

This equation is the form of a linear differential equation as: \(\frac{d y}{d x}+P y=Q \quad\left(\text { where } P=\frac{1}{x \log x} \text { and } Q=\frac{2}{x^2}\right)\)

Now, I.F. = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{x \log x} d x}=\mathrm{e}^{\log (\log x)}=\log \mathrm{x}\)

The general solution of the given differential equation is given by the relation,

y(I.F .) = \(\int(Q \times \text { I.F. }) d x+C \Rightarrow y \log x=\int\left(\frac{2}{x^2} \log x\right) d x+C\)

Now, \(\int\left(\frac{2}{x^2} \log x\right) d x=2 \int\left(\log x \cdot \frac{1}{x^2}\right) d x\)

= \(2\left[\log x \cdot \int \frac{1}{x^2} d x-\int\left\{\frac{d}{d x}(\log x) \cdot \int \frac{1}{x^2} d x\right\} d x\right]\)

= \(2\left[\log x\left(-\frac{1}{x}\right)-\int\left(\frac{1}{x} \cdot\left(-\frac{1}{x}\right)\right) d x\right]=2\left[-\frac{\log x}{x}+\int \frac{1}{x^2} d x\right]\)

= \(2\left[-\frac{\log x}{x}-\frac{1}{x}\right]=-\frac{2}{x}(1+\log x)\)

Substituting the value of \(\int\left(\frac{2}{x^2} \log x\right) d x\) in equation (1), we get : \(y \log x=-\frac{2}{x}(1+\log x)+C\)

This is the required general solution of the given differential equation.

Question 8. \(\left(1+x^2\right) d y+2 x y d x=\cot x d x(x \neq 0)\)
Solution:

(1+ \(x^2\)) dy+2 x y dx =cot x d x

⇒ \(\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{\cot x}{1+x^2}\)

This equation is a linear differential equation of the form:

⇒ \(\frac{d y}{d x}+P y=Q\)(where P = \(\frac{2 x}{1+x^2}\) and Q = \(\frac{\cot x}{1+x^2}\))

Now, I.F. = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\log \left(1+\mathrm{x}^2\right)}=1+\mathrm{x}^2\)

The general solution of the given differential equation is given by the relation,

y(I.F.) = \(\int(Q \times \text { I.F. }) d x+C \Rightarrow y\left(1+x^2\right)=\int\left[\frac{\cot x}{1+x^2} \times\left(1+x^2\right)\right] d x+C\)

⇒ y\(\left(1+x^2\right)=\int \cot x d x+C \Rightarrow y\left(1+x^2\right)=\log |\sin x|+C\)

Question 9. \(x \frac{d y}{d x}+y-x+x y \cot x=0(x \neq 0)\)
Solution:

⇒ \(x \frac{d y}{d x}+y-x+x y \cot x=0 \Rightarrow x \frac{d y}{d x}+y(1+x \cot x)=x \Rightarrow \frac{d y}{d x}+\left(\frac{1}{x}+\cot x\right) y=1\)

This equation is a linear differential equation of the form:

⇒ \(\frac{d y}{d x}+P y=Q\) (where P = \(\frac{1}{x}+\cot x\) and Q= 1)

Now, I.F. \(=\mathrm{e}^{\int P d x}=\mathrm{e}^{\int\left(\frac{1}{x}+\cot x\right) d x}=e^{\log x+\log (\sin x)}=e^{\log (x \sin x)}=x \sin x\)

The general solution of the given differential equation is given by the relation,

y(I.F.) = \(\int(Q \times \text { I.F. }) d x+C\)

⇒ \(y(x \sin x)=\int(1 \times x \sin x) d x+C \Rightarrow y(x \sin x)=\int(x \sin x) d x+C \)

⇒ \(y(x \sin x)=x \int \sin x d x-\int\left[\frac{d}{d x}(x) \cdot \int \sin x d x\right]+C \)

⇒ \(y(x \sin x)=x(-\cos x)-\int 1 \cdot(-\cos x) d x+C\)

⇒ \(y(x \sin x)=-x \cos x+\sin x+C \Rightarrow y=\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{C}{x \sin x}\)

⇒ \(y=-\cot \cdot x+\frac{1}{x}+\frac{C}{x \sin x}\) (which is the required solution)

Question 10. \((x+y) \frac{d y}{d x}=1\)
Solution:

⇒ \((x+y) \frac{d y}{d x}=1 \Rightarrow \frac{d y}{d x}=\frac{1}{x+y} \Rightarrow \frac{d x}{d y}=x+y \Rightarrow \frac{d x}{d y}-x=y\)

This is a linear differential equation of the form: \(\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{P}_1 \mathrm{x}=\mathrm{Q}_1\) (where \(\mathrm{P}_1=-1\) and \(\mathrm{Q}_1=\mathrm{y}\))

Now, I.F. = \(\mathrm{e}^{\int P_1 d y}=\mathrm{e}^{\int-\mathrm{dy}}=\mathrm{e}^{-\mathrm{y}}\)

The general solution of the given differential equation is given by the relation,

x(I.F.) = \(\int\left(Q_1 \times I . F .\right) d y+C\)

⇒ \(x^{-y}=\int\left(y \cdot e^{-y}\right) d y+C \Rightarrow x e^{-y}=y \cdot \int e^{-y} d y-\int\left[\frac{d}{d y}(y) \int e^{-y} d y\right] d y+C\)

⇒ \(x^{-y}=y\left(-e^{-y}\right)-\int\left(-e^{-y}\right) d y+C \Rightarrow \mathrm{xe}^{-y}=-y e^{-y}+\int e^{-y} d y+C\)

⇒ \(x^{-y}=-y e^{-y}-e^{-y}+C \Rightarrow x=-y-1+C e^z\)

⇒ \(x+y+1=C e^y\) (which is the required solution)

Question 11. \(y d x+\left(x-y^2\right) d y=0\)
Solution:

⇒ \(y d x+\left(x-y^2\right) d y=0 \Rightarrow y d x=\left(y^2-x\right) d y \Rightarrow \frac{d x}{d y}\)

= \(\frac{y^2-x}{y}=y-\frac{x}{y} \Rightarrow \frac{d x}{d y}+\frac{x}{y}=y\)

This is a linear differential equation of the form:

⇒ \(\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{P}_1 \mathrm{x}=\mathrm{Q}_1\) (where \( \mathrm{P}_1=\frac{1}{\mathrm{y}}\) and \(\mathrm{Q}_1=\mathrm{y}\))

Now. I.F. = \(\mathrm{e}^{\int {p_1} d y}=\mathrm{e}^{\int \frac{1}{y} d y}=\mathrm{e}^{\log y}=y\)

The general solution of the given differential equation is given by the relation,

x( I.F.) = \(\int\left(Q_1 \times I . F .\right) d y+C \Rightarrow x y=\int(y \cdot y) d y+C \Rightarrow x y=\int y^2 d y+C\)

⇒ xy = \(\frac{y^3}{3}+C \Rightarrow x=\frac{y^2}{3}+\frac{C}{y}\)

Question 12. \(\left(x+3 y^2\right) \frac{d y}{d x}=y(y>0)\)
Solution:

⇒ \(\left(x+3 y^2\right) \frac{d y}{d x}=y \Rightarrow \frac{d y}{d x}=\frac{y}{x+3 y^2} \Rightarrow \frac{d x}{d y}\)

= \(\frac{x+3 y^2}{y}=\frac{x}{y}+3 y \Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y\)

This is a linear differential equation of the form:

⇒ \(\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{P}_1 \mathrm{x}=\mathrm{Q}_1\) (where \(\mathrm{P}_1=-\frac{1}{\mathrm{y}}\) and \(\mathrm{Q}_{\mathbf{l}}=3 \mathrm{y}\))

Now, I.F. = \(\mathrm{e}^{\int P_1 d y}=\mathrm{e}^{-\int \frac{d y}{y}}=\mathrm{e}^{-\log y}=\mathrm{e}^{\log \left(\frac{1}{y}\right)}=\frac{1}{y}\)

The general solution of the given differential equation is given by the relation,

x(I .F.) = \(\int\left(Q_1 \times \text { I.F. }\right) d y+C\)

⇒ \(x \times \frac{1}{y}=\int\left(3 y \times \frac{1}{y}\right) d y+C \Rightarrow \frac{x}{y}=3 y+C \Rightarrow x=3 y^2+C y\)

For Each Of The Differential Equations, Find The Particular Solution Satisfying The Given Conditions:

Question 13. \(\frac{d y}{d x}+2 y \tan x=\sin x ; y=0\) when \(x=\frac{\pi}{3}\)
Solution:

The given differential equation is \(\frac{d y}{d x}+2 y \tan x=\sin x\)

This is a linear equation of the form: \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}\)(where \(\mathrm{P}=2 \tan \mathrm{x}\) and \(\mathrm{Q}=\sin \mathrm{x}\))

Now, I.F. = \(\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int 2 \mathrm{sin} x d x}=\mathrm{e}^{2 \log |\sec x|}=\mathrm{e}^{\log \left(\sec ^2 \mathrm{x}\right)}=\sec ^2 \mathrm{x}\)

The general solution of the given differential equation is given by the relation,

y(I.F) = \(\int(Q \times \text { I.F. }) d x+C\)

⇒ \(y\left(\sec ^2 x\right)=\int\left(\sin x \cdot \sec ^2 x\right) d x+C \Rightarrow y \sec ^2 x=\int(\sec x \cdot \tan x) d x+C\)

⇒ \(y \sec ^2 x=\sec x+C\)

Put, y=0 and x \(=\frac{\pi}{3}\), in eq.(1)

Therefore, \(0 \times \sec ^2 \frac{\pi}{3}=\sec \frac{\pi}{3}+\mathrm{C} \Rightarrow 0=2+\mathrm{C} \Rightarrow \mathrm{C}=-2\)

Substituting C=-2 in equation (1), we get: \(y \sec ^2 x=\sec x-2 \Rightarrow y=\cos x-2 \cos ^2 x\)

Hence, the required solution of the given differential equation is \(y=\cos x-2 \cos ^2 x\).

Question 14. \(\left(1+x^2\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^2} ; y=0\) when x=1
Solution:

⇒ \(\left(1+x^2\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^2} \Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{1}{\left(1+x^2\right)^2}\)

This is a linear differential equation of the form: \(\frac{d y}{d x}+P y=Q\) (where P = \(\frac{2 x}{1+x^2}\) and Q = \(\frac{1}{\left(1+x^2\right)^2}\))

Now, I.F. \(=\mathrm{e}^{\int P \mathrm{Pdx}}=\mathrm{e}^{\int \frac{7 \mathrm{xdx}}{1+\mathrm{x}^2}}=\mathrm{e}^{\log \left(1+\mathrm{x}^2\right)}=1+\mathrm{x}^2\)

The general solution of the given differential equation is given by the relation,

y(I.F.) = \(\int(Q \times I . F .) d x+C\)

⇒ \(y\left(1+x^2\right)=\int\left[\frac{1}{\left(1+x^2\right)^2} \cdot\left(1+x^2\right)\right] d x+C\)

⇒ \(y\left(1+x^2\right)=\int \frac{1}{1+x^2} d x+C \Rightarrow y\left(1+x^2\right)=\tan ^{-1} x+C\)….(1)

Put y=0 and x=1 in eq.(1), we get

0 = \(\tan ^{-1} 1+C \Rightarrow C=-\frac{\pi}{4}\)

Substituting C = \(-\frac{\pi}{4}\) in equation (1), we get : \(y\left(1+x^2\right)=\tan ^{-1} x-\frac{\pi}{4}\)

This is the required general solution of the given differential equation.

Question 15. \(\frac{\mathrm{dy}}{\mathrm{dx}}-3 \mathrm{y} \cot \mathrm{x}=\sin 2 \mathrm{x} ; \mathrm{y}=2\) when \(\mathrm{x}=\frac{\pi}{2}\)
Solution:

The given differential equation is \(\frac{d y}{d x}-3 y \cot x=\sin 2 x\)

This is a linear differential equation of the form: \(\frac{d y}{d x}+P y=Q\) (where P=-3cot x and Q=sin 2 x)

Now, I.F. = \(\mathrm{e}^{\int P \mathrm{Pdx}}=\mathrm{e}^{-3 \int \cot x d x}=\mathrm{e}^{-3 \log |\sin x|}=\mathrm{e}^{\log \left|\frac{1}{\sin } \mathrm{x}\right|}=\frac{1}{\sin ^3 \mathrm{x}}\)

The general solution of the given differential equation is given by the relation,

y(I.F.) = \(\int(Q \times \text { I.F. }) d x+C\)

⇒ \(y \cdot \frac{1}{\sin ^3 x}=\int\left[\sin 2 x \cdot \frac{1}{\sin ^3 x}\right] d x+C \Rightarrow y \mathrm{cosec}^3 x=2 \int(\cot x \cos e c x) d x+C\)

⇒ \(y \mathrm{cosec}^3 x=-2 \mathrm{cosec} x+C \Rightarrow y=-\frac{2}{\mathrm{cosec}^2 x}+\frac{C}{\mathrm{cosec}^3 x}\)

⇒ y = \(-2 \sin ^2 x+C \sin ^3 x\)……(1)

Put y=2 and \(x=\frac{\pi}{2}\) in eq.(1) we get

2 = \(-2 \sin ^2 \frac{\pi}{2}+C \sin ^3 \frac{\pi}{2} \Rightarrow 2=-2+C \Rightarrow C=4\)

Substituting C=4 in equation (1), we get:

y = \(-2 \sin ^2 x+4 \sin ^3 x \Rightarrow y=4 \sin ^3 x-2 \sin ^2 x\)

Question 16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.
Solution:

Lel F (x, y) is the curve passing through the origin.

At point (x, y), the scope of the curve will be \(\frac{dy}{dx}\)

According to the given information: \(\frac{d y}{d x}=x+y \Rightarrow \frac{d y}{d x}-y=x\)

This is a linear differential equation of the form: \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}\) (where P=-1 and Q = x)

Now, I.F. = \(\mathrm{e}^{\int 1 \mathrm{~d} d \mathrm{x}}=\mathrm{e}^{\int(-1) d x}=\mathrm{e}^{-\mathrm{x}}\)

The general solution of the given differential equation is given by the relation,

y(I.F.) = \(\int(Q \times \text { I.F. }) d x+C \Rightarrow \mathrm{ye}^{-x}=\int x e^{-x} d x+C\)……(1)

Now, \(\int x e^{-x} d x=x \int e^{-x} d x-\int\left[\frac{d}{d x}(x) \cdot \int e^{-x} d x\right] d x\)

= \(-x e^{-x}-\int-e^{-x} d x=-x e^{-x}+\left(-e^{-x}\right)=-e^{-x}(x+1)\)

Substituting in equation (1), we get:

⇒ \(\mathrm{ye}^{-x}=-\mathrm{e}^{-3}(\mathrm{x}+1)+\mathrm{C}\)

⇒ \(\mathrm{y}=-(\mathrm{x}+1)+\mathrm{Ce}^x \Rightarrow \mathrm{x}+\mathrm{y}+1=\mathrm{Ce}^{\mathrm{x}}\)….(2)

The curve passes through the origin. So, put x=0, y=0, and equation (2) becomes:

0+0+1= \(\mathrm{Ce}^0\) C=1

Substituting C =1 in equation (2), we get: x+y+1=\(e^x\)

Hence, the required equation of the curve passing through the origin is \(x+y+1=e^x\)

Question 17. Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of ths slope of the tangent to the curve at that point by 5.
Solution:

Let F (x, y) be the curve, and let (x, y) be a point on the curve. The slope of the tangent to the curve at (x, y) is \(\frac{dy}{dx}\)

According to the given information: \(\frac{d y}{d x}+5=x+y \Rightarrow \frac{d y}{d x}-y=x-5\)

This is a linear differential equation of the form: \(\frac{d y}{d x}+P y=Q\) (where P=-1 and Q=x-5)

Now, I.F. = \(\mathrm{e}^{\int P d x}=\mathrm{e}^{\int(-1) d x}=\mathrm{e}^{-\mathrm{x}}\)

The general equation of the curve is given by the relation,

y(I.F.) = \(\int(Q \times I.F.) d x+C \Rightarrow y \cdot e^{-x}=\int(x-5) e^{-x} d x+C\)…..(1)

Now, \(\int(x-5) e^{-x} d x=(x-5) \int e^{-x} d x-\int\left[\frac{d}{d x}(x-5) \cdot \int e^{-x} d x\right] d x\)

= \((x-5)\left(-e^{-x}\right)-\int\left(-e^{-x}\right) d x=(5-x) e^{-x}+\left(-e^{-x}\right)=(4-x) e^{-x}\)

Therefore, equation (1) becomes: \(y^{-x x}=(4-x) e^{-x}+C\)

y = \(4-x+C e^x \Rightarrow x+y-4=C e^x\)…..(2)

The curve passes through the point (0,2)

Therefore, equation (2) becomes : \(0+2-4=\mathrm{Ce}^0 \Rightarrow-2=\mathrm{C} \Rightarrow \mathrm{C}=-2\)

Substituting C=-2 in equation (2), we get: \(\mathrm{x}+\mathrm{y}-4=-2 \mathrm{e}^{\mathrm{x}} \Rightarrow \mathrm{y}=4-\mathrm{x}-2 \mathrm{e}^{\mathrm{x}}\)

This is the required equation of the curve.

Choose The Correct Answer In The Following

Question 18. The integrating factor of the differential equation \(x \frac{d y}{d x}-y=2 x^2\) is

1. \(\mathrm{e}^{x}\)
2. \(\mathrm{e}^{-y}\)
3. \(\frac{1}{\mathrm{x}}\)
4. \(\mathrm{x}\)

Solution: 3. \(\frac{1}{\mathrm{x}}\)

The given differential equation is: \(x \frac{d y}{d x}-y=2 x^2 \Rightarrow \frac{d y}{d x}-\frac{y}{x}=2 x\)

This is a linear differential equation of the form: \(\frac{d y}{d x}+P y=Q\) (where \(P=-\frac{1}{x}\) and Q=2x)

The integrating factor (I.F.) is given by the relation, \(\mathrm{e}^{\int \mathrm{Pdx}}\)

∴ I.F. = \(\mathrm{e}^{\int-\frac{1}{x} d x}=\mathrm{e}^{-\log x}=\mathrm{e}^{\log \left[x^{-1}\right\}}=\mathrm{x}^{-1}=\frac{1}{\mathrm{x}}\)

Hence, the correct answer is 3.

Question 19. The integrating factor of the differential equation \(\left(1-y^2\right) \frac{d x}{d y}+y x=a y(-1<y<1)\) is

1. \(\frac{1}{y^2-1}\)
2. \(\frac{1}{\sqrt{y^2-1}}\)
3. \(\frac{1}{1-y^2}\)
4. \(\frac{1}{\sqrt{1-y^2}}\)

Solution: 4. \(\frac{1}{\sqrt{1-y^2}}\)

The given differential equation is: \(\left(1-y^2\right) \frac{d x}{d y}+y x=a y \Rightarrow \frac{d x}{d y}+\frac{y x}{1-y^2}=\frac{a y}{1-y^2}\)

This is a linear differential equation of the form: \(\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q}\) (where \(\mathrm{P}=\frac{\mathrm{y}}{1-\mathrm{y}^2}\) and \(\mathrm{Q}=\frac{\mathrm{ay}}{1-\mathrm{y}^2}\))

The integrating factor (I.F.) is given by the relation,

∴ I.F. = \(\mathrm{e}^{\int \text { P.dy }}=\mathrm{e}^{\int \frac{y}{1-y^2} d y}=\mathrm{e}^{-\frac{1}{2} \log \left(1-y^2\right)}=\mathrm{e}^{\log \left[\frac{1}{\sqrt{1-y^2}}\right]}=\frac{1}{\sqrt{1-y^2}}\)

Hence, the correct answer is 4.

Differential Equations Miscellaneous Exercise

Question 1. For each of the differential equations given below, indicate its order and degree (if defined).

1. \(\frac{d^2 y}{d x^2}+5 x\left(\frac{d y}{d x}\right)^2-6 y=\log x\)
2. \(\left(\frac{d y}{d x}\right)^3-4\left(\frac{d y}{d x}\right)^2+7 y=\sin x\)
3. \(\frac{\mathrm{d}^4 y}{\mathrm{dx}^4}-\sin \left(\frac{\mathrm{d}^3 \mathrm{y}}{\mathrm{dx}^3}\right)=0\)

Solution:

1. The differential equation is given as:

⇒ \(\frac{d^2 y}{d x^2}+5 x\left(\frac{d y}{d x}\right)^2-6 y=\log x \Rightarrow \frac{d^2 y}{d x^2}+5 x\left(\frac{d y}{d x}\right)^2-6 y-\log x=0\)

The highest order derivative present in the differential equation is \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\).

Thus, its order is two. The highest power raised to \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) is one. Hence, its degree is one.

2. The differential equation is given as:

⇒ \(\left(\frac{d y}{d x}\right)^3-4\left(\frac{d y}{d x}\right)^2+7 y=\sin x \Rightarrow\left(\frac{d y}{d x}\right)^3-4\left(\frac{d y}{d x}\right)^2+7 y-\sin x=0\)

The highest order derivative present in the differential equation is \(\frac{d y}{d x}\).

Thus, its order is one. The highest power raised to \(\frac{\mathrm{dy}}{\mathrm{dx}}\) is three. Hence, its degree is three.

3. The differential equation is given as: \(\frac{d^4 y}{d x^4}-\sin \left(\frac{d^3 y}{d x^3}\right)=0\)

The highest order derivative p-=resent in the differential equation, \(\frac{d^4 y}{d x^4}\). Thus, its order is four.

However, the given differential equation is not a polynomial equation in its derivatives,

Hence, its degree is not defined.

Question 2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

1. \(y=a e^x+b e^{-x}+x^2 \quad: \quad x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}-x y+x^2-2=0\)
2. \(\mathrm{y}=\mathrm{e}^{\mathrm{x}}(\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}) \quad: \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}-2 \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y}=0\)
3. \(\mathrm{y}=\mathrm{x} \sin 3 \mathrm{x} \quad: \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}+9 \mathrm{y}-6 \cos 3 \mathrm{x}=0\)
4. \(x^2=2 y^2 \log y \quad: \left(x^2+y^2\right) \frac{d y}{d x}-x y=0\)

Solution:

1. \(\mathrm{y}=a \mathrm{e}^{\mathrm{x}}+b \mathrm{e}^{-\mathrm{x}}+\mathrm{x}^2\)

Differentiating both sides with respect to x, we get:

⇒ \(\frac{d y}{d x}=a \frac{d}{d x}\left(e^x\right)+b \frac{d}{d x}\left(e^{-x}\right)+\frac{d}{d x}\left(x^2\right) \Rightarrow \frac{d y}{d x}=a e^x-b e^{-x}+2 x\)

Again, differentiating both sides with respect to x, we get: \(\frac{d^2 y}{d x^2}=a e^x+b e^{-x}+2\)

Now, on substituting the values of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) and \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) in the differential equation, we get:

L.H.S. = \(x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}-x y+x^2-2\)

= \(x\left(a e^x+b e^{-x}+2\right)+2\left(a e^x-b e^{-x}+2 x\right)-x\left(a e^x+b e^{-x}+x^2\right)+x^2-2\)

= \(\left(a x e^x+b x e^{-x}+2 x\right)+\left(2 a e^x-2 b e^{-x}+4 x\right)-\left(a x e^x+b x e^{-4}+x^3\right)+x^2-2\)

= \(2 \mathrm{ae}^{\mathrm{x}}-2 \mathrm{be}^{-\mathrm{x}}-\mathrm{x}^3+\mathrm{x}^2+6 \mathrm{x}-2 \neq 0\)

⇒ \({ L.H.S. } \neq \text { R.H.S. }\)

Hence, the given function is not a solution to the corresponding differential equation.

2. \(\mathrm{y}=\mathrm{e}^{\mathrm{x}}(\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x})=a \mathrm{e}^{\mathrm{x}} \cos \mathrm{x}+b \mathrm{e}^x \sin \mathrm{x}\)

Differentiating both sides with respect to x, we get:

⇒ \(\frac{d y}{d x}=a \cdot \frac{d}{d x}\left(e^x \cos x\right)+b \cdot \frac{d}{d x}\left(e^x \sin x\right)\)

⇒ \(\frac{d y}{d x}=a\left(e^x \cos x-e^x \sin x\right)+b \cdot\left(e^x \sin x+e^x \cos x\right)\)

⇒ \(\frac{d y}{d x}=(a+b) e^x \cos x+(b-a) e^x \sin x\)

Again, differentiating both sides with respect to x, we get:

⇒ \(\frac{d^2 y}{d x^2}=(a+b) \cdot \frac{d}{d x}\left(e^x \cos x\right)+(b-a) \frac{d}{d x}\left(e^x \sin x\right)\)

⇒ \(\frac{d^2 y}{d x^2}=(a+b) \cdot\left[e^x \cos x-e^x \sin x\right]+(b-a)\left[e^x \sin x+e^x \cos x\right]\)

⇒ \(\frac{d^2 y}{d x^2}=e^x[(a+b)(\cos x-\sin x)+(b-a)(\sin x+\cos x]\)

⇒ \(\frac{d^2 y}{d x^2}=e^x[a \cos x-a \sin x+b \cos x-b \sin x+b \sin x+b \cos x-a \sin x-a \cos x]\)

⇒ \(\frac{d^2 y}{d x^2}=\left[2 e^x(b \cos x-a \sin x)\right]\)

Now, on substituting the values of \(\frac{\mathrm{d}^2 y}{\mathrm{dx}^2}\) and \(\frac{\mathrm{dy}}{\mathrm{dx}}\) in the L.H.S. of the given differential equation, we get:

⇒ \(\frac{d^2 y}{d x^2}+ 2 \frac{d y}{d x}+2 y\)

= \(2 e^x(b \cos x-a \sin x)-2 e^x[(a+b) \cos x+(b-a) \sin x]+2 e^x(a \cos x+b \sin x)\)

= \(e^x\left[\begin{array}{l}
(2 b \cos x-2 a \sin x)-(2 a \cos x+2 b \cos x) \\
-(2 b \sin x-2 a \sin x)+(2 a \cos x+2 b \sin x)
\end{array}\right]\)

= \(e^2[(2 b-2 a-2 b+2 a) \cos x]+e^x[(-2 a-2 b+2 a+2 b) \sin x]=0\)

Hence, the given function is a solution of the corresponding differential equation.

3. \(\mathrm{y}=\mathrm{x} \sin 3 \mathrm{x}\)

Differentiating both sides with respect to x, we get: \(\frac{d y}{d x}=\frac{d}{d x}(x \sin 3 x)=\sin 3 x+x \cdot \cos 3 x \cdot 3 \Rightarrow \frac{d y}{d x}=\sin 3 x+3 x \cos 3 x\)

Again, differentiating both sides with respect to x, we get:

⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}(\sin 3 x)+3 \frac{d}{d x}(x \cos 3 x)\)

⇒ \(\frac{d^2 y}{d x^2}=3 \cos 3 x+3[\cos 3 x+x(-\sin 3 x) \cdot 3] \Rightarrow \frac{d^2 y}{d x^2}=6 \cos 3 x-9 x \sin 3 x\)

Substituting the value of \(\frac{\mathrm{d}^2 y}{\mathrm{dx}^2}\) in the L.H.S. of the given differential equation, we get:

⇒ \(\left.\frac{d^2 y}{d x^2}+9 y-6 \cos 3 x=(6 \cos 3 x-9 x \sin 3 x)+9 x \sin 3 x-6 \cos 3 x\right)=0\)

Hence, the given function is a solution of the corresponding differential equation.

4. \(x^2=2 y^2 \log y\)

Differentiating both sides with respect to x, we get:

2x = \(2 \cdot \frac{d}{d x}\left[y^2 \log y\right]\)

⇒ \(x=\left[2 y \cdot \log y \cdot \frac{d y}{d x}+y^2 \cdot \frac{1}{y} \cdot \frac{d y}{d x}\right] \Rightarrow x=\frac{d y}{d x}(2 y \log y+y)\)

⇒ \(\frac{d y}{d x}=\frac{x}{y(1+2 \log y)}\)

Substituting the value of \(\frac{d y}{d x}\) in the L.H.S. of the given differential equation, we get:

L.H.S. = \(\left(x^2+y^2\right) \frac{d y}{d x}-x y=\left(2 y^2 \log y+y^2\right) \cdot \frac{x}{y(1+2 \log y)}-x y\)

= \(y^2(1+2 \log y) \cdot \frac{x}{y(1+2 \log y)}-x y=x y-x y=0\)

Hence, the given function is a solution of the corresponding differential equation.

Question 3. Prove that \(x^2-y^2=c\left(x^2+y^2\right)^2\) is the general solution of differential equation \(\left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y\), where c is a parameter.
Solution:

⇒ \(\left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y \Rightarrow \frac{d y}{d x}=\frac{x^3-3 x y^2}{y^3-3 x^2 y}\)

This is a homogeneous differential equation. To simplify it, we need to make the substitution as:

⇒ \(\frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting the values of y and \(\frac{d y}{d x}\) in equation (1), we get:

v+x \(\frac{d v}{d x}=\frac{x^3-3 x(v x)^2}{(v x)^3-3 x^2(v x)}\)

⇒ v+x \(\frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v} \Rightarrow x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v}-v\)

⇒ x \(\frac{d v}{d x}=\frac{1-3 v^2-v\left(v^3-3 v\right)}{v^3-3 v}\)

⇒ \(x \frac{d v}{d x}=\frac{1-v^4}{v^3-3 v} \Rightarrow\left(\frac{v^3-3 v}{1-v^4}\right) d v=\frac{d x}{x}\)

⇒ \(\int\left(\frac{v^3-3 v}{1-v^4}\right) d v=\int \frac{d x}{x}\)….(2)

⇒ \(\int\left(\frac{v^3-3 v}{1-v^4}\right) d v=\log x+\log C^{\prime}\)

Now, \(\int\left(\frac{v^3-3 v}{1-v^4}\right) d v=\int \frac{v^3 d v}{1-v^4}-3 \int \frac{v d v}{1-v^4}\)

⇒ \(\int\left(\frac{v^3-3 v}{1-v^4}\right) d v=I_1-3 I_2\), where \(I_1=\int \frac{v^3 d v}{1-v^4}\) and \(I_2=\int \frac{v d v}{1-v^4}\)….(3)

Let \(1-v^4=t\)

∴ \(\frac{d}{d v}\left(1-v^4\right)=\frac{d t}{d v} \Rightarrow-4 v^3=\frac{d t}{d v} \Rightarrow v^3 d v=-\frac{d t}{4}\)

Now, \(I_1=\int \frac{-d t}{4 t}=-\frac{1}{4} \log t=-\frac{1}{4} \log \left(1-v^4\right)\)

And, \(I_2=\int \frac{v d v}{1-v^4}=\int \frac{v d v}{1-\left(v^2\right)^2}\)

Let \(v^2=p\)

⇒ \(\frac{d}{d v}\left(v^2\right)=\frac{d p}{d v} \Rightarrow 2 v=\frac{d p}{d v} \Rightarrow v d v=\frac{d p}{2}\)

⇒ \(I_2=\frac{1}{2} \int \frac{d p}{1-p^2}=\frac{1}{2 \times 2} \log \left|\frac{1+p}{1-p}\right|=\frac{1}{4} \log \left|\frac{1+v^2}{1-v^2}\right|\)

Substituting the values of \(\mathrm{I}_1\) and \(\mathrm{I}_2\) in equation (3), we get:

⇒ \(\int\left(\frac{v^3-3 v}{1-v^4}\right) d v=-\frac{1}{4} \log \left(1-v^4\right)-\frac{3}{4} \log \left|\frac{1+v^2}{1-v^2}\right|\)

Therefore, equation (2) becomes:

⇒ \(-\frac{1}{4} \log \left(1-v^4\right)-\frac{3}{4} \log \left|\frac{1+v^2}{1-v^2}\right|=\log x+\log C^{\prime}\)

⇒ \(-\frac{1}{4} \log \left[\left(1-v^4\right)\left(\frac{1+v^2}{1-v^2}\right)^3\right]=\log C^{\prime} x\)

⇒ \(\frac{\left(1+v^2\right)^4}{\left(1-v^2\right)^2}=\left(C^{\prime} x\right)^{-1}\)

⇒ \(\frac{\left(1+\frac{y^2}{x^2}\right)^4}{\left(1-\frac{y^2}{x^2}\right)^2}=\frac{1}{\left(C^{\prime}\right)^4 x^4}\)

⇒ \(\frac{\left(x^2+y^2\right)^4}{x^4\left(x^2-y^2\right)^2}=\frac{1}{\left(C^{\prime}\right)^4 x^4}\)

⇒ \(\left(x^2-y^2\right)^2=\left(C^{\prime}\right)^4\left(x^2+y^2\right)^4\)

⇒ \(\left(x^2-y^2\right)=\left(C^{\prime}\right)^2\left(x^2+y^2\right)^2\)

⇒ \(x^2-y^2=c\left(x^2+y^2\right)^2\), where c = \(\left(C^{\prime}\right)^2\)

Hence, the given result is proved.

Question 4. Find the general solution of the differential equation \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0\)
Solution:

⇒ \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0 \Rightarrow \frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}}\)

⇒ \(\frac{d y}{\sqrt{1-y^2}}=\frac{-d x}{\sqrt{1-x^2}} \Rightarrow \int \frac{d y}{\sqrt{1-y^2}}=\int \frac{-d x}{\sqrt{1-x^2}}\)

⇒ \(\sin ^{-1} y=-\sin ^{-1} x+C \Rightarrow \sin ^{-1} x+\sin ^{-1} y=C\)

Question 5. Show that the general solution of the differential equation \(\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0\) is given by (x+y+1)=A(1-x-y-2 x y), where A is a parameter
Solution:

⇒ \(\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0 \Rightarrow \frac{d y}{d x}=-\frac{\left(y^2+y+1\right)}{x^2+x+1}\)

⇒ \(\frac{d y}{y^2+y+1}=\frac{-d x}{x^2+x+1} \Rightarrow \frac{d y}{y^2+y+1}+\frac{d x}{x^2+x+1}=0\)

⇒ \(\int \frac{d y}{y^2+y+1}+\int \frac{d x}{x^2+x+1}=C\)

⇒ \(\int \frac{d y}{\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\int \frac{d x}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=C\)

⇒ \(\frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]+\frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]=C\)

⇒ \(\tan ^{-1}\left[\frac{2 y+1}{\sqrt{3}}\right]+\tan ^{-1}\left[\frac{2 x+1}{\sqrt{3}}\right]=\frac{\sqrt{3 C}}{2}\)

⇒ \(\tan ^{-1}\left[\frac{\frac{2 y+1}{\sqrt{3}}+\frac{2 x+1}{\sqrt{3}}}{1-\frac{(2 y+1)}{\sqrt{3}} \frac{(2 x+1)}{\sqrt{3}}}\right]\)

= \(\frac{\sqrt{3} \mathrm{C}}{2}\)

⇒ \(\tan ^{-1}\left[\frac{\frac{2 x+2 y+2}{\sqrt{3}}}{1-\left(\frac{4 x y+2 x+2 y+1}{3}\right)}\right]=\frac{\sqrt{3} C}{2}\)

⇒ \(\tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{3-4 x y-2 x-2 y-1}\right]=\frac{\sqrt{3} C}{2}\)

⇒ \(\tan ^{-1}\left[\frac{\sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}\right]=\frac{\sqrt{3} C}{2}\)

⇒ \(\frac{\sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}=\tan \left(\frac{\sqrt{3} C}{2}\right)=B\) (where B = \(\tan \frac{\sqrt{3} C}{2}\))

⇒ x+y+1 = \(\frac{2 B}{\sqrt{3}}(1-x-y-2 x y)\)

⇒ \(\mathrm{x}+\mathrm{y}+\mathrm{I}=\mathrm{A}(1-\mathrm{x}-\mathrm{y}-2 \mathrm{xy})\) (where \(\mathrm{A}=\frac{2 \mathrm{~B}}{\sqrt{3}}\))

Hence, the given result is proved.

Question 6. Find the equation of the curve passing through the point \(\left(0, \frac{\pi}{4}\right)\) whose differential equation is, \(\sin x \cos y \mathrm{dx}+\cos \mathrm{x} \sin \mathrm{y} \mathrm{dy}=0\)
Solution:

The differential equation of the given curve is: sin x cos y d x+cos x sin y d y=0

⇒ \(\frac{\sin x \cos y d x+\cos x \sin y d y}{\cos x \cos y}=0 \Rightarrow \tan x d x+\tan y d y=0\)

Integrating both sides, we get \(\int \tan x d x+\int \tan y d y=\log C\)

⇒ log (sec x)+log (sec y)=log C

⇒ log (sec x sec y)=log C

⇒ sec x sec y=C….(1)

The curve passes through point \(\left(0, \frac{\pi}{4}\right)\)

∴ \(1 \times \sqrt{2}=\mathrm{C} \Rightarrow \mathrm{C}=\sqrt{2}\)

On substituting \(\mathrm{C}=\sqrt{2}\) in equation (1), we get:

⇒ \(\sec \mathrm{x} \cdot \sec \mathrm{y}=\sqrt{2}\)

⇒ \(\sec x \cdot \frac{1}{\cos y}=\sqrt{2} \Rightarrow \cos y=\frac{\sec x}{\sqrt{2}}\)

Hence, the required equation of the curve is \(\cos y=\frac{\sec x}{\sqrt{2}}\)

Question 7. Find the particular solution of the differential equation \(\left(1+e^{2 x}\right) d y+\left(1+y^2\right) e^x d x=0\), given that y=1 when x=0
Solution:

⇒ \(\left(1+e^{2 x}\right) d y+\left(1+y^2\right) e^x d x=0\)

⇒ \(\frac{d y}{1+y^2}+\frac{e^x d x}{1+e^{2 x}}=0 \Rightarrow \int \frac{d y}{1+y^2}+\int \frac{e^x d x}{1+e^{2 x}}=C\)

⇒ \(\tan ^{-1} y+\int \frac{e^x d x}{1+e^{2 x}}=C\)…..(1)

Let \(\mathrm{e}^{\mathrm{x}}=\mathrm{t} \Rightarrow \mathrm{e}^{2 \mathrm{x}}=\mathrm{t}^2\)

⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\mathrm{x}}\right)=\frac{\mathrm{dt}}{\mathrm{dx}}\)

⇒ \(\mathrm{e}^{\mathrm{x}}=\frac{\mathrm{dt}}{\mathrm{dx}} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\)

Substituting these values in equation (1), we get: \(\tan ^{-1} \mathrm{y}+\int \frac{\mathrm{dt}}{1+\mathrm{t}^2}=\mathrm{C}\)

⇒ \(\tan ^{-1} \mathrm{y}+\tan ^{-1} \mathrm{t}=\mathrm{C} \Rightarrow \tan ^{-1} \mathrm{y}+\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{C}\)….(2)

Now, y=1 at x=0.

Therefore, equation (2) becomes: \(\tan ^{-1} 1+\tan ^{-1} 1=\mathrm{C} \Rightarrow \frac{\pi}{4}+\frac{\pi}{4}=\mathrm{C} \Rightarrow \mathrm{C}=\frac{\pi}{2}\)

Substituting \(C=\frac{\pi}{2}\) in equation (2), we get: \(\tan ^{-1} y+\tan ^{-1}\left(e^x\right)=\frac{\pi}{2}\)

This is the required particular solution of the given differential equation.

Question 8. Solve the differential equation \(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^2\right) d y(y \neq 0)\)
Solution:

⇒ \(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^2\right) d y \Rightarrow y e^{\frac{x}{y}} \frac{d x}{d y}=x e^{\frac{x}{y}}+y^2\)

⇒ \(e^{\frac{x}{y}}\left[y \cdot \frac{d x}{d y}-x\right]=y^2\)…..(1)

Let \(\mathrm{e}^{\frac{\mathrm{x}}{y}}=\mathrm{z}\)

Differentiating it with respect to y, we get:

⇒ \(\frac{d}{d y}\left(e^{\frac{x}{y}}\right)=\frac{d z}{d y} \Rightarrow e^{\frac{x}{y}} \cdot \frac{d}{d y}\left(\frac{x}{y}\right)=\frac{d z}{d y}\)

⇒ \(e^{\frac{x}{y}} \cdot \frac{\left.y \cdot \frac{d x}{d y}-x\right]}{y^2}=\frac{d z}{d y}\)…..(2)

From equation (1) and equation (2), we get : \(\frac{\mathrm{dz}}{\mathrm{dy}}=1 \Rightarrow \mathrm{dz}=\mathrm{dy}\)

Integrating both sides, we get : \(\int d z=\int d y \Rightarrow z=y+C \Rightarrow e^{\frac{x}{y}}=y+C\)

Question 9. Find a particular solution of the differential equation (x-y)(d x+d y)=d x-d y, given that y=-1, when x=0(Hint : put x-y=t)
Solution:

(x-y)(d x+d y)=d x-d y

⇒ (x-y+1) d y=(1-x+y) d x

⇒ \(\frac{d y}{d x}=\frac{1-x+y}{x-y+1} \quad \Rightarrow \frac{d y}{d x}=\frac{1-(x-y)}{1+(x-y)}\)

Let x-y=t

⇒ \(\frac{d}{d x}(x-y)=\frac{d t}{d x} \Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}\)….(1)

⇒ \(1-\frac{d t}{d x}=\frac{d y}{d x}\)

Substituting the values of x-y and \(\frac{d y}{d x}\) in equation (1), we get :

⇒ \(1-\frac{d t}{d x}=\frac{1-t}{1+t}\)

⇒ \(\frac{d t}{d x}=1-\left(\frac{1-t}{1+t}\right) \Rightarrow \frac{d t}{d x}=\frac{(1+t)-(1-t)}{1+t} \Rightarrow \frac{d t}{d x}=\frac{2 t}{1+t}\)

⇒ \(\left(\frac{1+t}{t}\right) d t=2 d x \Rightarrow\left(1+\frac{1}{t}\right) d t=2 d x\)

Integrating both sides, we get: \(\int\left(1+\frac{1}{t}\right) d t=2 \int 1 \cdot d x\)

t + \(\log |t|=2 x+C \Rightarrow(x-y)+\log |x-y|=2 x+C\)

⇒ log |x-y|=x+y+C…..(2)

Now, y=-1 at x=0

Therefore, equation (2) becomes: log 1=0-1+C

⇒ C=1

Substituting C=1 in equation (2) we get: log |x-y|=x+y+1

This is the required particular solution of the given differential equation.

Question 10. Solve the differential equation \(\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1(x \neq 0)\)
Solution:

⇒ \(\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1 \Rightarrow \frac{d y}{d x}\)

= \(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \Rightarrow \frac{d y}{d x}+\frac{y}{\sqrt{x}}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}\)

This equation is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\), where P = \(\frac{1}{\sqrt{x}}\) and Q = \(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}\)

Now, I.F. = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{1}{\sqrt{x}} \mathrm{dx}}=\mathrm{e}^{2 \sqrt{x}}\)

The general solution of the given differential equation is given by,

y(I.F.) = \(\int(\text { Q } \times \text { I.F. }) \mathrm{dx}+\mathrm{C}\)

⇒ \(\mathrm{ye}^{2 \sqrt{x}}=\int\left(\frac{\mathrm{e}^{-2 \sqrt{x}}}{\sqrt{\mathrm{x}}} \times \mathrm{e}^{2 \sqrt{\mathrm{x}}}\right) \mathrm{dx}+\mathrm{C}\)

⇒ \(\mathrm{ye}^{2 \sqrt{\mathrm{x}}}=\int \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}+\mathrm{C} \Rightarrow \mathrm{ye}^{2 \sqrt{x}}=2 \sqrt{\mathrm{x}}+\mathrm{C}\)

Question 11. Find a particular solution of the differential equation \(\frac{d y}{d x}+y \cot x=4 x \mathrm{cosec} x(x \neq 0)\), given that y=0 where \(x=\frac{\pi}{2}\).
Solution:

The given differential equation is : \(\frac{d y}{d x}+y \cot x=4 x \mathrm{cosec} x\)

This equation is a linear differential equation of the form \(\frac{d y}{d x}+P y=0\), where P = cot x and Q = 4x cosec x

Now, I.F, \(=e^{\int P d x}=e^{\int \cot x d x}=e^{\log \sin x x}=\sin x\)

The general solution of the given differential equation is given by, \(y(\mathrm{I} . \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) \mathrm{dx}+\mathrm{C}\)

⇒ \(\mathrm{y} \sin \mathrm{x}=\int(4 \mathrm{x} \mathrm{cosec} \mathrm{x} \cdot \sin \mathrm{x}) \mathrm{dx}+\mathrm{C}\)

⇒ \(\mathrm{y} \sin \mathrm{x}=4 \int \mathrm{xdx}+\mathrm{C}\)

⇒ \(\mathrm{y} \sin \mathrm{x}=4 \cdot \frac{\mathrm{x}^2}{2}+\mathrm{C} \Rightarrow \mathrm{y} \sin \mathrm{x}=2 \mathrm{x}^2+\mathrm{C}\)…..(1)

Now, y=0 at x = \(\frac{\pi}{2}\)

Therefore, equation (1) becomes : 0 = \(2 \times \frac{\pi^2}{4}+\mathrm{C} \Rightarrow \mathrm{C}=-\frac{\pi^2}{2}\)

Substituting C = \(-\frac{\pi^2}{2}\) in equation (1), we get: \(y \sin x=2 x^2-\frac{\pi^2}{2}\)

This is the required particular solution of the given differential equation.

Question 12. Find a particular solution of the differential equation \((x+1) \frac{d y}{d x}=2 e^{-y}-1\), given that y=0 when x=0
Solution:

(x+1) \(\frac{d y}{d x}=2 e^{-y}-1 \Rightarrow \frac{d y}{2 e^{-y}-1}=\frac{d x}{x+1} \Rightarrow \frac{e^y d y}{2-e^y}=\frac{d x}{x+1}\)

Integrating both sides, we get : \(\int \frac{e^y d y}{2-e^y}=\int \frac{d x}{x+1}\)

⇒ \(\int \frac{\mathrm{e}^y \mathrm{dy}}{2-\mathrm{e}^y}=\log |\mathrm{x}+1|+\log \mathrm{C}\)…..(1)

Let \(2-e^{y}=\mathrm{t} \Rightarrow e^y d y=-d t\)

Substituting this value in equation (1), we get : \(-\int \frac{\mathrm{dt}}{\mathrm{t}}=\log |\mathrm{x}+1|+\log \mathrm{C}\)

⇒ \(-\log |\mathrm{t}|=\log |\mathrm{C}(\mathrm{x}+1)| \Rightarrow-\log \left|2-\mathrm{e}^{\mathrm{y}}\right|\)

= \(\log |\mathrm{C}(\mathrm{x}+1)| \Rightarrow \frac{1}{2-\mathrm{e}^{\mathrm{y}}}=\mathrm{C}(\mathrm{x}+1)\)

⇒ \(2-\mathrm{e}^{\mathrm{y}}=\frac{1}{\mathrm{C}(\mathrm{x}+1)}\)……(2)

Now, at x=0 and y=0, equation (2) becomes : 2-1 = \(\frac{1}{C}\)

C=1

Substituting C=1 in equation (2), we get : \(2-e^y=\frac{1}{x+1}\)

⇒ \(e^y=2-\frac{1}{x+1} \Rightarrow e^y=\frac{2 x+2-1}{x+1} \Rightarrow e^y=\frac{2 x+1}{x+1}\)

⇒ \(y=\log \left|\frac{2 x+1}{x+1}\right|,(x \neq-1)\)

This is the required particular solution of the given differential equation.

Choose The Correct Answer

Question 13. The general solution of the differential equation \(\frac{y d x-x d y}{y}=0\) is

1. xy=C
2. \(x=C y^2\)
3. y=Cx
4. \(\mathrm{y}=\mathrm{Cx}^2\)

Solution: 3. y=Cx

The given differential equation is: \(\frac{y \mathrm{dx}-\mathrm{x} d \mathrm{y}}{\mathrm{y}}=0\)

⇒ \(\frac{1}{\mathrm{x}} \mathrm{dx}-\frac{1}{\mathrm{y}} \mathrm{dy}=0\)

Integrating both sides, we get : \(\int \frac{1}{x} d x-\int \frac{1}{y} d y=0 \Rightarrow \log |x|-\log |y|=\log k \Rightarrow \log \left|\frac{x}{y}\right|=\log k\)

⇒ \(\frac{x}{y}=k \Rightarrow y=\frac{1}{k} x \Rightarrow y=C x\) where c = \(\frac{1}{k}\)

Hence, the correct answer is (3).

Question 14. The general solution of a differential equation of the type \(\frac{d x}{d y}+P_1 x=Q_1\) is

1. \(y \mathrm{e}^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)
2. \(y \cdot e^{\int P_1 d x}=\int\left(Q_1 e^{\int P_1{d x}}\right) d y+C\)
3. \(x e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)
4. \(x e^{\int P_1 d x}=\int\left(Q_1 e^{\int P_1 d x}\right) d y+C\)

Solution: 3. \(x e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)

The integrating factor of the given differential equation \(\frac{d x}{d y}+P_1 x=Q_1\) is \(e^{\int P_1 d y}\).

The general solution of the differential equation is given by,

x(I .F.) = \(\int\left(Q_1 \times \text { I.F. }\right) d y+C \Rightarrow x \cdot e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)

Hence, the correct answer is (3).

Question 15. The general solution of the differential equation \(e^x d y+\left(y e^x+2 x\right) d x=0\) is

1. \(x e^y+x^2=C\)
2. \(x e^y+y^2=C\)
3. \(y \mathrm{e}^{\mathrm{x}}+\mathrm{x}^2=\mathrm{C}\)
4. \(\mathrm{ye}^y+\mathrm{x}^2=\mathrm{C}\)

Solution: 3. \(y \mathrm{e}^{\mathrm{x}}+\mathrm{x}^2=\mathrm{C}\)

The given differential equation is: \(e^x d y+\left(y e^x+2 x\right) d x=0\)

⇒ \(e^x \frac{d y}{d x}+y e^x+2 x=0 \quad \Rightarrow \quad \frac{d y}{d x}+y=-2 x e^{-x}\)

This is a linear differential equation of the form

⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}\) where P =1 and Q=-2 \(\mathrm{x} \mathrm{e}^{-\mathrm{t}}\)

Now, I.F. = \(\mathrm{e}^{\int P d x}=\mathrm{e}^{\int d x}=\mathrm{e}^\pi\)

The general solution of the given differential equation is given by, \(y(\text { I.F. })=\int(\mathrm{Q} \times \text { I.F. }) \mathrm{dx}+C\)

⇒ \(\mathrm{ye}^{\mathrm{x}}=\int\left(-2 \mathrm{xe}^{-\mathrm{x}} \cdot \mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}+\mathrm{C}\)

⇒ \(\mathrm{ye}^{\mathrm{x}}=-\int 2 \mathrm{xdx}+\mathrm{C}\)

⇒ \(\mathrm{ye}^{\mathrm{x}}=-\mathrm{x}^2+\mathrm{C}\)

⇒ \(\mathrm{ye}^{\mathrm{x}}+\mathrm{x}^2=\mathrm{C}\)

Hence, the correct answer is 3.