#### Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 29

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

Hint: To solve this equation we use  $\frac{d y}{d x}+P y=Q$  where $P, Q$  are constants.

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

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

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

$If$ of differential equation is

\begin{aligned} &\text { If }=e^{-\int \frac{2 x}{1+x^{2}} d x} \quad\left[1+x^{2}=t, 2 x d x=d t\right] \\ & \end{aligned}

$=e^{-\int \frac{d t}{t}} \\$

$=e^{-\ln t} \\$

$=e^{\ln \left(t^{-1}\right)} \\$

$=t^{-1} \\$

$=\frac{1}{t}$

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

$y \text { If }=\int \text { QIf } d x+C \\$

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

$=\int \frac{x^{2}+1+1}{x^{2}+1} d x+C \\$

$=\int \frac{x^{2}+1}{x^{2}+1} d x+\int \frac{1}{x^{2}+1} d x+C \\$

$=\int d x+\tan ^{-1} x+C$

\begin{aligned} &=x+\tan ^{-1} x+C \\ & \end{aligned}

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