Answer
The solutions of the differential equation :
$$
x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0, \quad \mu(x, y)=1 / x y^{3}
$$
are given implicitly by
$$
x^{2}-y^{-2}+2\ln |y| =c
$$
where $ c $ is an arbitrary constant
Work Step by Step
$$
x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0, \quad \mu(x, y)=1 / x y^{3} \quad \quad (i)
$$
this equation can be written as
$$
\left(x^{2} y^{3} \right) d x+\left(x \left(1+y^{2}\right) \right)d y=0
$$
Comparing this Equation with the differential form:
$$
M(x,y) d x+N(x,y) d y=0
$$
we observe that
$$
\begin{aligned}M(x,y) &= x^{2} y^{3}
\\ N(x,y) &=x \left(1+y^{2}\right) \end{aligned}
$$
By calculating $M_{y}$ and $N_{x}$ , we find that
$$
\begin{aligned} M_{y}(x, y) &=3x^{2} y^{2}
\\ N_{x}(x, y) & = 1+y^{2} \end{aligned}
$$
we obtain that
$$
M_{y}(x, y) \neq N_{x}(x, y)
$$
so the given equation is not exact.
Multiplying Eq. (i) by this integrating factor $ \mu(x, y)=1 / x y^{3}$, we obtain
$$
x d x+\left( \frac{\left(1+y^{2}\right)}{y^{3}} \right)d y=0 \quad \quad (ii)
$$
The latter equation is exact, since
$$
M_{y}(x, y)=0=N_{x}(x, y)
$$
the equation (ii) is exact. Thus there is a $\psi (x, y)$ such that
$$
\begin{aligned} \psi_{x}(x, y) &=M(x,y) =x ,
\\ \psi_{y}(x, y) &=N(x,y) = \frac{\left(1+y^{2}\right)}{y^{3}} \end{aligned} \quad \quad (iii)
$$
Integrating the first of these equations with respect to $x$ , we obtain
$$
\psi(x, y)= \frac{1}{2}x^{2}+h(y) \quad \quad (*)
$$
where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (iii), we compute $\psi_{y}(x, y) $ from Eq. (*) and set it equal to $N$, obtaining
$$
\psi_{y}(x, y)= h^{\prime}(y)= \frac{\left(1+y^{2}\right)}{y^{3}}
$$
Thus
$$
\begin{split}
h(y) &=\int{\left( \frac{\left(1+y^{2}\right)}{y^{3}} \right) }dy \\
&=\int{\left( \frac{\left(1+y^{2}\right)}{y^{3}} \right) }dy
\\
&=\int{\frac{1}{y^{3}}}dy+\int{\frac{1}{y}}dy
\\
&=-\frac{1}{2}y^{-2}+\ln |y|
\end{split}
$$
The constant of integration can be omitted since any solution of the preceding differential equation is satisfactory.
Now substituting for h(y) in Eq. (iii) gives
$$
\psi(x, y)=\frac{1}{2}x^{2}-\frac{1}{2}y^{-2}+\ln |y|
$$
Hence solutions of Eq. (i) are given implicitly by
$$
\psi(x, y) =x^{2}-y^{-2}+2\ln |y| =c
$$
where $ c $ is an arbitrary constant