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Provide solution for RD Sharma class 12 chapter 21 Differential Equations exercise 21.6 question 4

Answers (1)

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Answer: $x+\cot y+y=C$

Given: $\frac{d y}{d x}=\frac{1-\cos 2 y}{1+\cos 2 y}$

Hint: Use trigonometric identities to simplify and integrate

Solution

$\frac{d y}{d x}=\frac{1-\cos 2 y}{1+\cos 2 y}$ $\left[\cos 2 x=1-\sin ^{2} x=2 \cos ^{2} x-1\right]$

$\frac{d y}{d x}=\frac{2 \sin ^{2} y}{2 \cos ^{2} y}$ $\left[\frac{\sin x}{\cos x}=\tan x\right]$

$\frac{d y}{d x}=\tan ^{2} y$

$\frac{1}{\tan ^{2} y} d y=d x$ $\left[\frac{1}{\tan x}=\cot x\right]$

$\int \cot ^{2} y d y=\int d x$ $\left[\cot ^{2} y=\operatorname{cosec}^{2} y-1\right]$

$\int\left(\operatorname{cosec}^{2} y-1\right) d y=\int d x$ $\left[\int \operatorname{cosec}^{2} y d y=-\cot y+c\right]$

$\begin{aligned} &-\cot y-y+C=x \\ &-\cot y-y-x=-C \\ &C=\cot y+y+x \end{aligned}$

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