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  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 60 textbook solution.

Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 60 textbook solution.

Answers (1)

Answer : x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+c

Hint               : You must know the rules of solving differential equation and integration

Given             :   \left(1+y^{2}\right) d x=\left(\tan ^{-1} y-x\right) d y

Solution         : \left(1+y^{2}\right) d x=\left(\tan ^{-1} y-x\right) d y

                                    \begin{aligned} \frac{d x}{d y} &=\frac{\tan ^{-1} y-x}{1+y^{2}} \\ \frac{d x}{d y}+\frac{x}{1+y^{2}} &=\frac{\tan ^{-1} y}{1+y^{2}} \end{aligned}

Comparing with, \frac{dx}{dy}+Px=Q,

\begin{aligned} &P=\frac{1}{1+y^{2}}, Q=\frac{\tan ^{-1} y}{1+y^{2}} \\ &\text { Now } \quad \text { I. } F=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y} \end{aligned}

So, the solution is ,

\begin{aligned} x \times I . F &=\int(I . F \times Q) d y+c \\ x e^{\tan ^{-1} y} &=\int \frac{\tan ^{-1} y}{1+y^{2}} \times e^{\tan ^{-1} y} d y+c \\ x e^{\tan ^{-1} y} &=I+C \end{aligned}

Now, I=\int \frac{\tan ^{-1} y}{1+y^{2}} \times e^{\tan ^{-1} y} d y

\begin{aligned} &\text { Put }_{\mu} \text { tan }^{-1} y=t \text { and differentiate, }\\ &\begin{gathered} \quad \frac{1}{1+y^{2}} d y=d t \\ I=\int e^{t} t d t \end{gathered} \end{aligned}

Integrating by parts,

\begin{aligned} &I=t \times \int e^{t} d t-\int\left\{\frac{d t}{d t} \int e^{t} d t\right\} d t \\ &I=t e^{t}-\int e^{t} d t \\ &I=t e^{t}-e^{t} \end{aligned}

\begin{aligned} &I=\tan ^{-1} y e^{\tan ^{-1} y}-e^{\tan ^{-1} y} \\ &I=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right) \end{aligned}

By putting value of I

We get the required solution,

x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+c

Posted by

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