Question

Integrating factor of the differential equation \((1-y²) \frac{dx}{dy} + xy = ay\) is:

• A. \(\frac{1}{1-y^2}\)
• B. \(\frac{1}{\sqrt{y^2-1}}\)
• C. \(\frac{1}{{y^2-1}}\)
• D. \(\frac{1}{\sqrt{1-y^2}}\)

Answer 1

The Correct Option is D

The given differential equation is: \((1-y²) \frac{dx}{dy} + xy = ay\). To find the integrating factor, we need to express it in the form \(\frac{dx}{dy} + P(y)x = Q(y)\). We can rewrite the equation as:

\[\frac{dx}{dy} + \frac{xy}{1-y^2} = \frac{ay}{1-y^2}\]

Here, we have \(P(y) = \frac{y}{1-y^2}\). An integrating factor \(\mu(y)\) for an equation of this form is given by:

\[\mu(y) = e^{\int P(y)dy}\]

Let's compute the integral:

\[\int P(y)dy = \int \frac{y}{1-y^2}dy\]

Using the substitution \(u = 1-y^2\), \(du = -2y \, dy\), we have:

\[\int \frac{y}{1-y^2}dy = -\frac{1}{2} \int \frac{1}{u}du = -\frac{1}{2} \ln|u| = -\frac{1}{2} \ln|1-y^2|\]

Thus,

\[\mu(y) = e^{-\frac{1}{2} \ln|1-y^2|} = (1-y^2)^{-\frac{1}{2}}\]

So, the integrating factor is:

\[\mu(y) = \frac{1}{\sqrt{1-y^2}}\]