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  • Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 28 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 28 Maths Textbook Solution.

Answers (1)

Answer: x\tan x-y\tan |\frac{\sec x}{\sec y}|+C

Hint: Use the formula of \int u\: v\: dx

Given: x\cos ^{2}y\: dx=y\: \cos ^{2}x\: dy

Solution:x\cos ^{2}y\: dx=y\: \cos ^{2}x\: dy

\begin{aligned} &\Rightarrow \quad \frac{x}{\cos ^{2} x} d x=\frac{y}{\cos ^{2} y} d y \\ &\Rightarrow \quad y \sec ^{2} y d y=x \sec ^{2} x d x \quad\left(\because \frac{1}{\cos x}=\sec x\right) \\ &\Rightarrow \int y \sec ^{2} y d y=\int x \sec ^{2} x d x \quad(\because \text { integrating both sides }) \\ &\Rightarrow \quad y \int \sec ^{2} y d y-\int \tan y d y=x \int \sec ^{2} x d x-\int \tan x d x \\ &\quad\left[\because \int u v d x=u \int v d x-\int\left(\frac{d}{d x} u \int v d x\right) d x\right. \\ &\Rightarrow \quad y \tan y-\log |\sec y|=x \tan x-\log |\sec x|+C \\ &\Rightarrow \quad x \tan x-y \tan y=\log |\sec x|-\log |\sec y|+C \\ &\Rightarrow \quad x \tan x-y \tan y=\log \left|\frac{\sec x}{\sec y}\right|+C \quad[\because \log a-\log b=\log (a / b)] \end{aligned}

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