$\begingroup$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!
$\endgroup$13 Answers
$\begingroup$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.
$\endgroup$1$\begingroup$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: $$\boxed {y(x)=x^2 + C_1 \ln x+C_2}$$
$\endgroup$2$\begingroup$$$x y'' + y' = 4 x$$
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.
$\endgroup$1
