Linear Second Order ODE/y'' + y = 0/Proof 5

Proof
Taking Laplace transforms, we have:


 * $\laptrans {y'' + y} = \laptrans 0$

From Laplace Transform of Constant Mapping, we have:


 * $\laptrans 0 = 0$

We also have:

So:


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

Giving:


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

So:

Setting $C_1 = \map {y'} 0$ and $C_2 = \map y 0$ gives the result.