#### Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 29 maths textbook solution.

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

Hint : To solve this equation we use $\frac{dy}{dx}+Py=Q$ 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}

$I\: f$ 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] \\ &=e^{-\int \frac{d t}{t}} \\ &=e^{-\ln t} \end{aligned}

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

\begin{aligned} &y I f=\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 \end{aligned}

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