Question

The integrating factor of the differential equation $\left(1-y^2\right)\frac{dx}{dy}+yx=ay\left(-1<y<1\right)$ 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

$\left(\mathrm{d}\right)\frac{1}{\sqrt{1-y^2}}$


$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(1-y^2\right)\frac{dx}{dy}+yx=ay \\ \frac{dx}{dy}+\frac{y}{1-y^2}x=\frac{ay}{1-y^2} \\ \mathrm{Comparing}\mathrm{with}\frac{dx}{dy}+Px=Q,\mathrm{we}\mathrm{get} \\ P=\frac{y}{1-y^2} \\ Q=\frac{ay}{1-y^2} \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \frac{y}{1-y^2}dy} \\ =e^{-\frac{1}{2}\int \frac{-2y}{1-y^2}dy} \\ =e^{-\frac{1}{2}\log \left|1-y^2\right|} \\ =e^{\log \left|\frac{1}{{\sqrt{1-y}}^2}\right|} \\ =\frac{1}{{\sqrt{1-y}}^2} \end{gathered}$$