#### Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (x) maths textbook solution.

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

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

Hint : Using $\int \sec ^{2} x d x$

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

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

This is a first order linear differential equation of the form

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

The integrating factor $If$  of this differential equation is

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

Hence, the solution of different equation is

\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 \end{aligned}

\begin{aligned} &=x e^{y}=\int \sec ^{2} y d y+C \; \; \; \; \; \; \quad\left[e^{-y} e^{y}=e^{-y+y}=e^{0}=1\right] \\ &=x e^{y}=\tan y+C \; \; \; \; \; \; \; \; \; \; \quad\left[\int \sec ^{2} x d x=\tan x+C\right] \end{aligned}