Find the general solution of the differential equation $xy'' + y' = 4x$ I began by rewriting this differential equation as:
$$ \frac { x \frac {d}{dx} w(x) + w(x) } {x} = 4$$
This has the form $y' + P(x)y = Q(x)$ and so $$ \frac {dw} {w} = -P(x)dx$$
Integrating this yields $w= \frac {C}{x}$ and thus $$ \frac {d}{dx}C(x) = 4x$$ so $C(x) = 2x^2 + C$.
But I'm not sure how to proceed with the rest of the solution. Any guidance is greatly appreciated!
 A: I can't follow your steps.
Instead, note that $xy'' + y' = (xy')'$, and the equation $$(xy')' = 4x$$ can be solved for $y$ by integration and basic algebra.
A: 
$$x y'' + y' = 4 x$$

The rewrite in the question statement is not correct, but rather it replaces $y'$ with $y$, giving a different differential equation.
One can set $u := y'$, on the other hand and so write the equation as a first order one,
$$x u' + u = 4 x ,$$
and apply the standard technique of looking for an integrating factor.
Hint Alternatively, notice that the left-hand side of the equation can already be written as $$(xy')' = 4 x ,$$
(i.e., we need not introduce an integrating factor), and integrating gives
$$x y ' = 2 x^2 + C,$$
which reduces the problem to rearranging and computing an antiderivative.
A: Your solution is not that wrong :
$$xy'' + y' = 4x$$
$$xw' + w = 4x$$
Solve the homogeneous equation:
$$xw'=-w$$
$$\ln w =-\ln x +C$$
$$\implies w=\dfrac {C(x)}{x}$$
Substitute this in the original DE:
$$x(\dfrac {C(x)}{x})'+\dfrac {C(x)}{x}=4x$$
$$x(\dfrac {C'(x)x-C(x)}{x^2})+\dfrac {C(x)}{x}=4x$$
$$C'(x)=4x$$
Integrate:
$$C(x)=2x^2+C$$
Therefore:
$$W=\dfrac {2x^2+C}{x}$$
$$W=2x +\dfrac Cx$$
Since $w=y'$ we have
$$y'=2x +\dfrac Cx$$
Integrate:
$$\boxed {y(x)=x^2 + C_1 \ln x+C_2}$$
