#### Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 67 Subquestion (i) textbook solution.

Answer :  $y\left(1+x^{2}\right)=\tan ^{-1} x-\frac{\pi}{4}$

Hint : Use the formula $\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c$

Given : $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}} ; y=0 \text { when } x=1$

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

Divide by $1+x^{2}$

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

This is a linear differential equation of the form

Here, $P=\frac{2 x}{1+x^{2}} \quad ; Q=\frac{1}{\left(1+x^{2}\right)^{2}}$

The integrating factor of this differential equation is

$\text { I.F }=e^{\int P d x}=e^{\int \frac{2 x}{1+x^{2}} d x}$

We have,      \begin{aligned} &\int \frac{2 x}{1+x^{2}}=\log \left|1+x^{2}\right|+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d x}{x}=\log |x|+c\right] \\ \end{aligned}

Therefore, \begin{aligned} &\text { I. } F=e^{\log \left|1+x^{2}\right|}=1+x^{2} \\ \end{aligned}

\begin{aligned} &\text { I. } F=1+x^{2} \end{aligned}

Hence the solution of the differential equation is

$\begin{gathered} y(I . F)=\int(Q \times I . F) d x+c \\ y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)^{2}}\left(1+x^{2}\right) d x+c \\ \Rightarrow y\left(1+x^{2}\right)=\int \frac{1}{1+x^{2}} d x+c \end{gathered}$                     ...(i)

We know , $\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c$

Therefore, $y\left(1+x^{2}\right)=\tan ^{-1} x+c$                                      ....(ii)

Now $y=0$ when $x=1$

Therefore,

\begin{aligned} 0\left(1+1^{2}\right) &=\tan ^{-1}(1)+c \\ \Rightarrow 0 &=\frac{\pi}{4}+c \Rightarrow c=-\frac{\pi}{4} \end{aligned}

By (ii)

$y\left(1+x^{2}\right)=\tan ^{-1} x-\frac{\pi}{4}$