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

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Answer : y=x \sin ^{-1} x+\sqrt{1-x^{2}}+c

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

Given              : \frac{d y}{d x}=\sin ^{-1} x

Solution         :  \frac{d y}{d x}=\sin ^{-1} x

                          \frac{d y}{d x}=\sin ^{-1} x\; dx

Integrating  both sides

                    \int d y=\int 1 \times \sin ^{-1} x d x

Integrating by parts,

 Put               \begin{aligned} \int d y &=\sin ^{-1} x \int 1 d x-\int\left[\frac{d}{d x}\left(\sin ^{-1} x\right) \int 1 d x\right] d x \\ y &=x \sin ^{-1} x-\int \frac{x}{\sqrt{1-x^{2}}} d x \\ t^{2} &=1-x^{2}, \text { and differentiate we get } \end{aligned}

                             \begin{gathered} 2 t d t=-2 x d x \\ \Rightarrow-t d t=x d x \end{gathered}


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

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