Question

For the differential equation, find the general solution:

${\displaystyle \frac{dy}{dx} ={\sin }^{-1}x}$

Answer 1

We have ${\displaystyle \frac{dy}{dx}={\sin }^{-1}x}$

⇒ dy = sin^-1 x dx                   ...(1)

Integrating (1) both sides, we get

${\displaystyle \int dy=\int {\sin }^{-1}xdx}$

⇒ ${\displaystyle y={\sin }^{-1}x\int 1dx-\int (\frac{d}{dx}\left({\sin }^{-1}x\right)\int 1dx) dx}$

⇒ ${\displaystyle y=x{\sin }^{-1}x-\int \frac{x}{\sqrt{1-x^2}} dx}$

⇒ ${\displaystyle y=x{\sin }^{-1}x+\frac{1}{2}\int \frac{(-2x)dx}{\sqrt{1-x^2}}}$

⇒ ${\displaystyle y=x{\sin }^{-1}x+\frac{1}{2}\int \frac{1}{\sqrt{t}} dt}$      

[Putting 1 - x² = t ⇒ -2x dx = dt]

⇒ ${\displaystyle y=x{\sin }^{-1}x+\frac{1}{2}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C}$

⇒ ${\displaystyle y=x{\sin }^{-1}x+\sqrt{t}+C}$

⇒ ${\displaystyle y=x{\sin }^{-1}x+\sqrt{1+x^2}+C}$

Which is the required solution.