The solution of the given differential equation $\sqrt{1+2y}dx+\sqrt{1-2x}dy=0$ is:

$\sqrt{1+2y}=\sqrt{1-2x}+C$

$-\sqrt{1-2x}dy=\sqrt{1+2y}dx$

$⇒\frac{dy}{\sqrt{1+2y}}=\frac{-dx}{\sqrt{1-2x}}$

Integrating both side.

$∫\frac{dy}{\sqrt{1+2y}}=-∫\frac{dx}{\sqrt{1-2x}}+C$

$\frac{(1+2y)^{\frac{-1}{2}+1}}{2.(\frac{-1}{2}+1)}=-\frac{(1+2x)^{\frac{-1}{2}+1}}{(-2)(\frac{-1}{2}+1)}+C$

 $\sqrt{1+2y}=\sqrt{1-2x}+C$