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Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation Excercise 21.10 Question 26

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Answer:  x=(\tan y+C) e^{-y}

Hint: To solve this equation we use  where  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 \\ & \end{aligned}

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

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} \\ & \end{aligned}

=e^{\int 1 d y} \\

\int d y=y+C \\

=I f=e^{y}                                        \quad\left[e^{\log x}=x\right]

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

=x e^{y}=\int e^{-y} \sec ^{2} y e^{y} d y+C \\

=x e^{y}=\int \sec ^{2} y d y+C \\

=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}


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