The expression which is the general solution of the differential equation dydx+x1−x2y=xy√ is

[a] Divide the equation by y√, we get$y-12dydx+x1-x2y12=x$Put y12=z⇒12y−12dydx=dzdx$2dzdx+x1-x2z=x\Rightarrow dzdx+\left(12x1-x2\right)z=x2$I. F. e∫12[x1−x2]dx=e−14log(1−x2)=(1−x2)14$Thesolution$is z(1−x2)−14=∫x2(1−x)−14dx+c$\Rightarrow z\left(1-x2\right)-14=12\int \left(1-x2\right)-12xdx+c$⇒y√(1−x2)14=12(−12)(1−x2)3/43/4+c$\Rightarrow y\surd \left(1-x2\right)14=-13\left(1-x2\right)\frac{3}{2}+c$⇒y√+13(1−x2)=c(1−x2)14