Question

For the differential equation in given question find the general solution.​​​​

$\frac{dy}{dx}=si{n}^{-1}x$

Answer 1

Given, $\frac{dy}{dx}=si{n}^{-1}x$
On separating the variables, we get $dy=si{n}^{-1}xdx$
On integrating , we get $\int dy=\int si{n}^{-1}xdx$
$y=si{n}^{-1}x\int 1dx-\int \left[\frac{d}{dx}\right(si{n}^{-1}x)\int 1dx]dx$ (using integration by parts)
$\Rightarrow y=xsi{n}^{-1}x-\int \left[\frac{x}{\sqrt{1-x^2}}\right]dxLet1-x^2=t\Rightarrow -2x=\frac{dt}{dx}=dx=\frac{dt}{-2x}\therefore y=si{n}^{-1}x+\int \frac{x}{\sqrt{t}}\frac{dt}{2x}\Rightarrow y=xsi{n}^{-1}x+\frac{1}{2}.\frac{t^{-1/2+1}}{-1/2+1}+C\Rightarrow y=xsi{n}^{-1}x+\frac{2}{2}\sqrt{t}+C\Rightarrow y=xsi{n}^{-1}x+\sqrt{1-x^2}+C$
which is the required general solution.