Question

Solve :

${\sin }^{-1}\left(\frac{dy}{dx}\right)=x+y$

Answer 1

$si{n}^{-1}\left(\frac{dy}{dx}\right)=x+y$

$\frac{dy}{dx}=\sin (x+y)$

Let $x+y=v$. Then,

$1+\frac{dy}{dx}=\frac{dv}{dx}$

$\frac{dy}{dx}=\frac{dv}{dx}-1$

Therefore,

$\frac{dv}{dx}-1=\sin v$

$\frac{dv}{dx}=1+\sin v$

$\frac{dv}{1+\sin v}=dx$

$\int \frac{1}{1+\sin v}=\int dx$

$\int dx=\int \frac{1-\sin v}{1-si{n}^{2}v}dv$

$\int dx=\int \frac{1-\sin v}{co{s}^{2}v}dv$

$\int dx=\int (se{c}^{2}v-\tan v\sec v)dv$

$x=\tan v-\sec v+C$

$x=\tan (x+y)-\sec (x+y)+C$, which is the required solution.