# Second-Order Differential Equation

$$xy'' + 2y' = 12x^2$$

Solve this by substituting $u = y'$

When I do this I get $u = y' = \frac{4x^3 + C}{2x}$

How do I continue?

Thanks

• separating variables equation $y' =\frac{4x^3 + C}{2x}\Rightarrow dy=\frac{4x^3 + C}{2x}dx$ – R.N Oct 23 '15 at 9:25
• And what am I supposed to do with that information? – John Snoe Oct 23 '15 at 9:32
• I'm not sure your solution for $u$ is correct. – Matthew Leingang Oct 23 '15 at 9:33

Letting $u=y'$ gives an an equation for $u$: $$x u' + 2u = 12 x^2 \implies u' + \frac{2}{x} u = 12x$$ This differential equation is linear with integrating factor $$p = e^{\int \frac{2}{x}\,dx} = e^{2 \ln x} = x^2$$ Multiplying through we have \begin{align*} x^2 u' + 2x u &= 12x^3 \\ \implies (x^2u)' &= 12x^3 \\ \implies x^2 u &= 3x^4+ C \\ \implies u &= 3x^2 + \frac{C}{x^2} \end{align*} You can check that $u$ satisfies the desired equation, for \begin{align*} u' &= 6x - \frac{2C}{x^3} \\ xu' &= 6x^2 - \frac{2C}{x^2} \\ 2u &= 6x^2 + \frac{2C}{x^2} \\ \end{align*} The sum of the RHS of the last two equations is $12x^2$.
But you wanted $y$, not $u$. No problem; we know $$y' = 3x^2 + Cx^{-2} \implies y = x^3 - C x^{-1} + D = x^3 - \frac{C}{x} + D$$ You can check this $y$ function satisfies your original DE.