Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x/Proof 2

Proof
Taking Laplace transforms:


 * $\laptrans {y'' + 4 y} = \laptrans {8 x^2 - 4 x}$

We have:

We also have:

So:


 * $\paren {s^2 + 4} \laptrans y = s \map y 0 + \map {y'} 0 + \dfrac {16 - 4 s} {s^3}$

Giving:

So:

Setting $C_1 = \dfrac {\paren {\map {y'} 0 + 1} } 2$ and $C_2 = \map y 0 + 1$ gives the result.