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Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (x) Maths Textbook Solution.

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Answer:  \begin{aligned} &x e^{y}=\tan y+C \\ & \end{aligned}

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

Hint: Using   \int \sec ^{2} x d x

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

 \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 \\ &=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}

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