The solution of the given differential equation $\sqrt{1+2y}dx+\sqrt{1-2x}dy=0$ is:
Answer & explanation
Correct answer: option 2
$-\sqrt{1-2x}dy=\sqrt{1+2y}dx$
$⇒\frac{dy}{\sqrt{1+2y}}=\frac{-dx}{\sqrt{1-2x}}$
Integrating both side.
$∫\frac{dy}{\sqrt{1+2y}}=-∫\frac{dx}{\sqrt{1-2x}}+C$
$\frac{(1+2y)^{\frac{-1}{2}+1}}{2.(\frac{-1}{2}+1)}=-\frac{(1+2x)^{\frac{-1}{2}+1}}{(-2)(\frac{-1}{2}+1)}+C$
$\sqrt{1+2y}=\sqrt{1-2x}+C$