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Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 64 Subquestion (ii) textbook solution.

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Answer : y=2 \sin (x+c)

Hint               : You must know the rules of solving differential equation and integration

Given            :    \frac{d y}{d x}=\sqrt{4-y^{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad,-2<y<2

Solution        :    \frac{d y}{d x}=\sqrt{4-y^{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad,-2<y<2

                     \frac{1}{\sqrt{4-y^{2}}}dy=dx

Integrating both sides,

            \begin{aligned} \int \frac{1}{\sqrt{4-y^{2}}} d y &=\int d x \\ \sin ^{-1} \frac{y}{2} &=x+c \\ \frac{y}{2} &=\sin (x+c) \\ y &=2 \sin (x+c) \end{aligned}

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