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Provide solution for RD Sharma class 12 chapter 21 Differential Equations exercise 21.6 question 3

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Answer: x+\cot y=C

Given: \frac{d y}{d x}=\sin ^{2} y

Hint: You must know trigonometric identities

Solution:  \frac{d y}{d x}=\sin ^{2} y

                \begin{aligned} &\frac{1}{\sin ^{2} y} d y=d x\\ &\operatorname{cosec}^{2} y d y=d x \quad \text { [applying integration] }\\ &\int \operatorname{cosec}^{2} y d y=\int d x \quad\: \; \; \; \; \; \; \; \; \; \left[\int \operatorname{cosec}^{2} y d y=-\operatorname{coty}+c\right]\\ &-\cot y=x+C\\ &x+\cot y=C \end{aligned}

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