The solution of the given differential equation \(\sqrt{1+2y}\,dx+\sqrt{1-2x}\,dy=0\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{1+2y}=\sqrt{1-2x}+C\)
The variables separate: dividing by \(\sqrt{1+2y}\,\sqrt{1-2x}\),
\( \displaystyle \frac{dy}{\sqrt{1+2y}} = -\frac{dx}{\sqrt{1-2x}} \)
Integrating both sides, using \(\int (ax+b)^{-1/2}\,dx = \frac{2}{a}\sqrt{ax+b}\):
\( \displaystyle \int (1+2y)^{-1/2}\,dy = \sqrt{1+2y} \)
\( \displaystyle -\int (1-2x)^{-1/2}\,dx = -\frac{2}{-2}\sqrt{1-2x} = \sqrt{1-2x} \)
So the solution is
\( \sqrt{1+2y} = \sqrt{1-2x} + C \)