Question

For the differential equation find the general solution of
$\frac{dy}{dx}={\sin }^{-1}x$

Answer 1

Given differential equation is

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

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

Integrating both sides
We get,

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

Using Integrating by parts:

$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int \left[f^{\prime }\right(x)\int g(x\left)dx\right]dx$

Take $f\left(x\right)={\sin }^{-1}x\text{and}g\left(x\right)=1$

Putting $\left(a\right)$ in $\left(i\right)$ then we get,

$y=x.{\sin }^{-1}x-\int \frac{-du}{2\sqrt{u}}$

$y=x.{\sin }^{-1}x+\frac{1}{2}\int \frac{-1}{u^2}dt$

$y=x.{\sin }^{-1}x+\frac{1}{2}\frac{u^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}+c$

$y=x.{\sin }^{-1}x+\sqrt{u}+c$

Now, putting the value of $u=1-x^2$ then we get,

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

Final Answer:
Hence, the required general solution is

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