Question

The integrating factor (IF) of the differential equation
$(1-y^2)\frac{dy}{dx}+yx=ay(-1<y<1)$ is
(a)$\frac{1}{y^2-1}$
(b)$\frac{1}{\sqrt{y^2-1}}$
(c)$\frac{1}{1-y^2}$
(d)$\frac{1}{\sqrt{1-y^2}}$

Answer 1

The given differential equation is $(1-y^2)\frac{dy}{dx}+yx=ay$, -1<y<1 (on dividing by $(1-y^2)$ both sides in the above equation)
$\Rightarrow \frac{dy}{dx}+\left(\frac{y}{1-y^2}\right)x=\frac{ay}{1-y^2}Here,P=\frac{y}{1-y^2},Q=\frac{ay}{1-y^2}andIF=e^{\int \frac{y}{1-y^2}dy}\Rightarrow IF=e^{-\frac{1}{2}\int \frac{-2y}{1-y^2}dy}\Rightarrow IF=e^{-\frac{1}{2}log(1-y^2)}\Rightarrow IF=e^{log(1-y^2{)}^{-1/2}}=(1-y^2{)}^{-1/2}=\frac{1}{\sqrt{1-y^2}}$
Hence, the correct option is (d).