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

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

Answers (1)

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}

 

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