Question

Solve the differential equation :

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

Answer 1

Given : ${\sin }^{-1}\left(\begin{array}{c}\frac{dy}{dx}\end{array}\right)=x+y$
$\frac{dy}{dx}=\sin (x+y)$ .............(i)

Put $v=x+y$, we get
$\frac{dv}{dx}=1+\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$

From (i), we have

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

Putting the value of $x+y$ and $\frac{dy}{dx}$, we get
$\frac{dv}{dx}-1=\sin v$
$\Rightarrow \frac{dv}{dx}=\sin v+1$
$\Rightarrow dx=\frac{1}{1+\sin v}dv$

On integrating both sides,

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

$x=\int \frac{1}{1+\sin v}\times \frac{1-\sin v}{1-\sin v}dv$

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

$=\int \frac{1-\sin v}{{\cos }^{2}v}dv$

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

$\therefore x=\tan v-\sec v+c$.