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

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}

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

Therefore,

\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}