Question

The intergrating factor of the differential equation $(1-y^2)\frac{dx}{dy}+yx=ay(-1<y<1)$ 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 correct option is D $\frac{1}{\sqrt{1-y^2}}$

For finding Integrating factor of given Differential equation we reduce it to a standard form:

$\frac{dx}{dy}+Px=Q$

Integrating factor for standard form: $I=e^{\int Pdy}$

$Given$ $e{q}^n=(1-y^2)\frac{dx}{dy}+yx=ay$ $[-1<y<1]$

$\Rightarrow \frac{dx}{dy}+\left(\frac{y}{1-y^2}\right)x=\frac{ay}{1-y^2}$

Now our differential equation is in standard form:

Here, $P=\frac{y}{1-y^2}$

so, $I=e^{\int \left(\frac{ydy}{1-y^2}\right)}$

Now,

$\int \frac{ydy}{1-y^2}\Rightarrow \int \left(\frac{-1}{2}\right)\left(\frac{-2ydy}{1-y^2}\right)$

$\Rightarrow \left(\frac{-1}{2}\right)\int \frac{-2ydy}{1-y^2}$

As, $\frac{d}{dy}(1-y^2)=-2ydy$ we get:

$\therefore$ $\int \frac{ydy}{1-y^2}$ $\Rightarrow \left(\frac{-1}{2}\right)ln(1-y^2)$

So,

$I=e^{\left(\frac{-1}{2}\right)ln(1-y^2)}$

$\Rightarrow e^{ln\left(\frac{1}{\sqrt{1-y^2}}\right)}\Rightarrow \frac{1}{\sqrt{1-y^2}}$ [$\because$ $a^{lnx}=x^{lna}$]

$\therefore I=\frac{1}{\sqrt{1-y^2}}$

Hence, Option (D) is correct