Linear Second Order ODE/y'' - y' - 6 y = exp -x/Proof 3

Proof
Taking Laplace transforms:


 * $\laptrans {y'' - y' - 6 y} = \laptrans {e^{-x} }$

We have:

We also have:

So:


 * $\paren {s^2 - s - 6} \laptrans y = s \map y 0 + \paren {\map {y'} 0 - \map y 0} + \dfrac 1 {s + 1}$

Giving:

So:

Setting:


 * $C_1 = \dfrac {8 \map y 0 + 4 \map {y'} 0 + 1} {20}$
 * $C_2 = \dfrac {3 \map y 0 - \map {y'} 0 + 1} 5$

gives the result.