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