The integrating factor of the differential equation \(\left(1-y^2\right)\frac{dx}{dy}+yx=ay\left(-1<y<1\right)\)
Answer & explanation
Correct answer: option 4
Convert into the form \(\frac{dy}{dx}+P(x)y=Q(x)\) then \(I.F=\int P(x)dx\)
\(\left(1-y^2\right)\frac{dx}{dy}+yx=ay\left(-1<y<1\right)\)
$\frac{dx}{dy} = \frac{y}{1-y^2} x = a \frac{y}{1-y^2} $
which is an exact DE of the form $\frac{dx}{dy}+P(y)x=Q(y)$ then \(I.F=\int P(y)dy\)
$ \text{Integrating factor is } e^{\int_{p(y) dy}} = \frac{1}{\sqrt{1-y^2}}$