Question
Find the integrating factor of the differential equation
$$
\frac{dy}{dx}
+
\frac{x}{1-x^2}y
=
x\sqrt{y}
$$
Solution
The given differential equation is
$$
\frac{dy}{dx}
+
\frac{x}{1-x^2}y
=
x\sqrt{y}
$$
Since the equation contains the term $\sqrt{y}$, it is not linear.
Let
$$
v=\sqrt{y}
$$
Then,
$$
y=v^2
$$
Differentiating,
$$
\frac{dy}{dx}
=
2v\frac{dv}{dx}
$$
Substitute these values into the given equation:
$$
2v\frac{dv}{dx}
+
\frac{x}{1-x^2}v^2
=
xv
$$
Divide both sides by $2v$:
$$
\frac{dv}{dx}
+
\frac{x}{2(1-x^2)}v
=
\frac{x}{2}
$$
This is now a linear differential equation of the form
$$
\frac{dv}{dx}+P(x)v=Q(x)
$$
where
$$
P(x)=
\frac{x}{2(1-x^2)}.
$$
The integrating factor is
$$
\text{I.F.}
=
e^{\int P(x)\,dx}
=
e^{\int
\frac{x}{2(1-x^2)}dx}
$$
Let
$$
u=1-x^2
$$
Then,
$$
du=-2x\,dx
$$
Therefore,
$$
\int
\frac{x}{2(1-x^2)}dx
=
-\frac14
\int
\frac{du}{u}
$$
Hence,
$$
=
-\frac14\ln|1-x^2|
$$
Therefore,
$$
\text{I.F.}
=
e^{-\frac14\ln(1-x^2)}
$$
Using the identity
$$
e^{\ln a}=a,
$$
we obtain
$$
\boxed{
\text{I.F.}
=
(1-x^2)^{-1/4}
}
$$
Hence, the correct answer is
$$
\boxed{(1-x^2)^{-1/4}}
$$
which corresponds to Option (B).