Question

Can ${\sin }^{-1}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=x+y$ be solved using the variable separable method?(yes/no)
Ans :

Answer 1

There are some differential equations which on the face look like can’t be solved using variable separable. One has to make proper substitutions to reduce it into a form where variables can be separated.
In this question we will put $x+y=v$
$x+y=v$
$1+\frac{dy}{dx}=\frac{dv}{dx}$

$\frac{dv}{dx}-1=\frac{dy}{dx}$ ...(1)
From the given equation ${\sin }^{-1}\left(dy/dx\right)=x+y$
$\frac{dy}{dx}=\sin (x+y)$
$=\sin \left(v\right)$ ...(2)
So from (1) and (2)
$\frac{dv}{dx}-1=\sin v$
$\Rightarrow \frac{dv}{[1+\sin v]}=dx$
$\Rightarrow \frac{dv}{[\sin v/2+\cos v/2{]}^2}=dx$
Taking ${\cos }^2\left(v/2\right)$ common from the denominator.
$\Rightarrow \frac{{\sec }^{2}v/2}{(1+\tan v/2{)}^2}dv=dx$
Now put 1+\tan v/2=t
$\frac{1}{2}{\sec }^{2}v/2dv=dt$
$\Rightarrow \frac{2}{t^2}dt=dx$
Integrating we will get
$\Rightarrow \frac{-2}{t}=x+cor\frac{-2}{1+\tan v/2}=(x+c)\Rightarrow (1+\tan v/2)(x+c)+2=0\Rightarrow (1+\tan \frac{(x+y)}{2})(x+c)+2=0$