#### Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 26 maths textbook solution.

Answer : $x=(\tan y+C) e^{-y}$

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

Give : $d x+x d y=e^{-y} \sec ^{2} y d y$

Solution :

\begin{aligned} & d x=e^{-y} \sec ^{2} y d y-x d y \\ &=\frac{d x}{d y}+x=e^{-y} \sec ^{2} y \end{aligned}

First order linear differential equation form

\begin{aligned} &=\frac{d x}{d y}=P x-Q \\ &P=1 \text { and } e^{-y} \sec ^{2} y \end{aligned}

If of differential equation is

\begin{aligned} &I f=e^{\int R d y} \\ &=e^{\int 1 d y} \\ &\int d y=y+C \\ &=I f=e^{y} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[e^{\log x}=x\right] \end{aligned}

\begin{aligned} &=x(I f)=\int Q I F d y+C \\ &=x e^{y}=\int e^{-y} \sec ^{2} y e^{y} d y+C \\ &=x e^{y}=\int \sec ^{2} y d y+C \end{aligned}

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