The correct option is A y=Ae2/3(2a−x)√x+a
Given dydx+xy√a+x=0⇒dyy=−xdx√a+x
Integrating both sides, ∫dyy=∫−x√x+adx
logy=−∫x+a−a√x+a=−∫√x+adx+∫a√x+adx⇒logy=−23(x+a)3/2+2a√x+a+logAy=Ae−2/3(x+a)3/2+2a√x+a=Ae[√x+a(−23(x+a)+2a)]=Ae[√x+a(−2x−2a+6a3)]=Ae[−2/3√x+a(x−2a)]
or y=Ae[2/e√x+a(2a−x)]