#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 20 textbook solution.

Answer : $2 y e^{\tan ^{-1} x}=e^{2 \tan ^{-1} y}+C$

Hint : To solve this equation we use $I \int f(x) d x$ formula.

Give : $\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$

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

\begin{aligned} &=\frac{d y}{d x}+\frac{y}{\left(1+x^{2}\right)}=\frac{e^{\tan ^{-1} x}}{1+x^{2}} \\ \end{aligned}                   .....(i)

\begin{aligned} &=\frac{d y}{d x}+P(x) y=Q x \\ \end{aligned}

\begin{aligned} &P=\frac{1}{1+x^{2}}, Q=\frac{e^{\tan ^{-1} x}}{1+x^{2}} \\ \end{aligned}

\begin{aligned} &I f=e^{\int P d x} \\ \end{aligned}

\begin{aligned} &=e^{\int \frac{1}{1+x^{2}} d x} \\ \end{aligned}

\begin{aligned} &=e^{\tan ^{-1} x} \end{aligned}                              .....(ii)

\begin{aligned} &=e^{\tan ^{-1} x} \frac{d y}{d x}+e^{\tan ^{-1} x} \frac{y}{1+x^{2}}=\frac{\left(e^{\tan ^{-1} x}\right)^{2}}{1+x^{2}} \\ \end{aligned}

\begin{aligned} &=\frac{d}{d x}\left[y e^{\tan ^{-1} x}\right]=\frac{\left(e^{\tan ^{-1} x}\right)^{2}}{1+x^{2}} \\ \end{aligned}

\begin{aligned} &=d\left(y e^{\tan ^{-1} x}\right)=\frac{e^{\tan ^{-1} x^{2}}}{1+x^{2}} d x \\ \end{aligned}

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