Solve the differential equation $ x^4y^3=xy' + y$ 
Using the change of the dependent variable $z = y^{−2}$, solve the differential equation:
$$xy' + y = x^4y^3$$.


My attempt:
$$xy' + y = x^4y^3 \tag 1$$
Now dividing $(1)$ by $y^2$
$$\frac{xy'}{y^2} + \frac{1}{y} = x^4y$$
Now put $z= \frac{1}{y^2}$,
$$\frac{dz}{dx} = \frac{-2}{y^3}\frac{dy}{dx}=\frac{-2}{y^3}y'$$
$$y'=\frac{-y^3dz}{2dx}$$
$$xy' z +\frac{1}{y} = x^4y$$
After that im not able to proceed further.
 A: So far, you have:
$z = \frac{1}{y^2}$
$z' = -\frac{2y'}{y^3}$
Your differential equation, after dividing through by $y^3$, becomes:
$\frac{xy'}{y^3}+\frac{1}{y^2} = x^4$
So, just use your substitution to turn the above into a linear differential equation in terms of $z$ and $z'$. Then, use an integrating factor to solve it completely for $z$. 
Once you do that, getting $y$ as a function of $x$ becomes an algebra problem.
I hope this assists you in solving the problem. 
A: It would have been easier to divide by $y^3$ initially instead of $y^2$. Also, instead of just replacing $\frac{1}{y^2}$ with $z$ in your first term, if you used your $y'=\frac{-y^3dz}{2dx}$ instead, you would then get
$$\begin{equation}\begin{aligned}
-\frac{xyz'}{2} +\frac{1}{y} & = x^4y \\
-\frac{xz'}{2} + \frac{1}{y^2} & = x^4 \\
-\frac{xz'}{2} + z & = x^4
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
I trust you can now proceed to solve this.
A: Hint.
$$
x(x y)^3=(x y)'
$$
and now making $z = xy$ we have
$$
x z^3= z'
$$
which is separable.
A: $$xy'+y=(xy)'$$ calls for the rewrite
$$x=\frac{(xy)'}{(xy)^3}.$$
This can be integrated straight away.
