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

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

Answers (1)

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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