#### Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 20

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

Hint: To solve this equation we use  $I\int f\left ( x \right )dx$  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} & \\ &\quad=\frac{d y}{d x}+\frac{y}{\left(1+x^{2}\right)}=\frac{e^{\tan ^{-1} x}}{1+x^{2}} \ldots(i) \\\\ &\quad=\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}} \\\\ &I f=e^{\int P d x} \\\\ &=e^{\int \frac{1}{1+x^{2}} d x} \\\\ &=e^{\tan ^{-1} x} \ldots(i i) \end{aligned}

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

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

$=y e^{\tan ^{-1} x}=\int \frac{e^{\tan ^{-1} x^{2}}}{1+x^{2}} d x+C$

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