Integrating factor for $xdx +(x-y^2)dy=0.$ How to find integrating factor for $xdx+(x-y^2)dy=0.$
Here $M=x$ and $N=x-y^2$. This equation is not exact. Clearly not seperable, homogeneous or linear(either in terms of $\frac{dy}{dx}$ or in terms of $\frac{dx}{dy}$). This is not even Bernoulli's equation. Also $\frac{1}{M}\left[\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right]=\frac 1 x,$ a function $x$, but since the rule requires this expression to be a function of $y$, this one is not applicable too.
 A: We need to solve $$\text{ODE [1]:} \quad \boxed{xdx+(x-y^{2})dy=0}$$
Now a very common technique is to swap the variables. Let $x \leftrightarrow y$, so we obtain $$\text{ODE [2]:}\quad \boxed{ydy+(y-x^{2})dx=0}$$Note that $\text{ODE [1]:}\overbrace{\equiv}^{x\leftrightarrow y} \text{ODE [2]}$. Now, we can see in $\color{blue}{[2]}$ that
\begin{eqnarray}
ydy+(y-x^{2})dx&=&0\\
 &\implies& ydy=(x^{2}-y)dx\\
 &\implies& yy'=x^{2}-y\\
 &\iff& yy'+y=x^{2}\\
 &\iff & y(y'+1)=x^{2}
 \end{eqnarray}
Now, we can see that $x=0 \implies y=0$. Therefore, we have the IPV
$$\text{(IPV):} \left\{\begin{aligned}
y(y'+1)=x^{2}\\
y(0)=0
\end{aligned}
\right.$$
We can solve IPV using power series. Let $$y=\sum_{k=1}^{+\infty}a_{k}x^{k}\implies y'=\sum_{k=2}^{+\infty}a_{k}kx^{k-1}$$So, we have to solve for $a_{k}$, $$\sum_{k=1}^{+\infty}a_{k}x^{k}\left( \sum_{k=2}^{+\infty}a_{k}kx^{k-1}+1 \right)=x^{2}$$I think the reader can do this last step, finally we obtain $$\boxed{y(x)=-x-\frac{x^{2}}{2}+\frac{x^{3}}{6}-\frac{5x^{4}}{48}+\frac{19x^{5}}{240}-\frac{287x^{6}}{4320}+\frac{3583x^{7}}{60480}-\cdots}$$
