$x^2 dy/dx + 2xy=y^{-3}$, find $y(x)$ The question states to start by substituting $u=y^{-2}$, however I am not sure how to deal with this as-as far as I can see there is no easy substitution. 
Are you meant to find that $y=(1/u)^{1/2}$ and then work with that(seems a little messy)? 
Thanks,
 A: You can write $(x^2 y)'=\frac{1}{y^3}=\frac{x^6}{(x^2 y)^3}$.
Multiplicated with $(x^2 y)^3$ we get $\frac{1}{4}((x^2 y)^4)'=x^6$.
Then integration for $x$ and then solving for $y$. 
A: Rewrite it as follows
$$
(2xy-y^{-3})dx+x^2dy=0
$$
Multiply it by $x^6y^3$, then
$$
(2x^7y^4-x^6)dx+x^8y^3dy=0
$$
Which is full differential. Hence:
$$
F=\int{x^8y^3dy}=\frac{x^8y^4}{4}+f(x)\\
\frac{dF}{dx}=2x^7y^4+\frac{df}{dx}=2x^7y^4-x^6\to\\
f=-\frac{x^7}{7}\to F=\frac{x^8y^4}{4}-\frac{x^7}{7}=C^*
$$
Which represent family of soutions:
$$
y^4=\frac{C}{x^8}+\frac{4}{7x}
$$
Edit
Consider substitution $u=y^{-2}\to du=-2y^{-3}dy$, then we have following equation:
$$
-\frac{x^2y^3}{2}\frac{du}{dx}+2xy=y^{-3}
$$
Divide both sides of equation by $y^3$
$$
-\frac{x^2}{2}\frac{du}{dx}+2xy^{-2}=y^{-6}\to
-\frac{x^2}{2}\frac{du}{dx}+2xu=u^3\to\\
(u^3-2xu)dx+\frac{x^2}{2}du=0
$$
Multiply equation by $u^{-3}x^6$, then we have full differential
$$
(x^6-2x^7u^{-2})dx+\frac{x^8u^{-3}}{2}du=0
$$
From which 
$$
F=\int\frac{x^8u^{-3}}{2}du=-\frac{x^8u^{-2}}{4}+f(x)\\
\frac{dF}{dx}=-2x^7u^{-2}+\frac{df}{dx}=x^6-2x^7u^{-2}\to\\
f=\frac{x^7}{7}\to F=\frac{x^7}{7}-\frac{x^8u^{-2}}{4}=C^*
$$
Which after inverse substitution leads to the same family of solutions
$$
y^4=\frac{C}{x^8}+\frac{4}{7x}
$$
A: The equation can be written as follows (divide both sides by $x^2$) 
$$\frac{dy}{dx} = \left(-\,\frac{2}{x}\right)\, y + \left(\frac{1}{x^2}\right) \, y^{-3}.$$
This is a special case of the following family of differential equations
$$\frac{dy}{dx} = a(x)\, y + b(x) \, y^n$$
The general method for solving these kind of equations is by adjusting $m$ of a substitution of the following form 
$$z(x) = y(x)^m\,\,\,\text{ with inverse } \,\,\,  y(x) = z(x)^{\frac{1}{m}}$$
or in short notations, skipping the argument $x$
$$z = y^m \,\,\,\text{ with inverse } \,\,\, y = z^{\frac{1}{m}}$$
The derivative with respect to $x$ of this substitution is 
$$\frac{dz}{dx} = m \, y^{m-1}\, \frac{dy}{dx}$$ 
Now, that implies the following chain of equalities
\begin{align}
\frac{dz}{dx} &= m \, y^{m-1}\, \frac{dy}{dx} =    m \, y^{m-1} \,\Big( a(x)\, y + b(x) \, y^n \Big) \\
&= m \, a(x) \, y^{m} + m\, b(x) \, y^{n + m -1} \\
&= m \, a(x) \, z + m \, b(x) \, y^{n + m -1}
\end{align} 
Now, notice that if we choose $m = 1- n$ then $n+m-1=0$ and the differential equations turns into the linear differential equation
$$\frac{dz}{dx} = m \, a(x) \, z + m \, b(x)$$
which is always solvable, because there is an explicit formula for its general solution. 
In your case, $n=-3$ and so $m = 4$. Then, a substitution that can work will be  $z = y^{4}$.   
