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

Theorem

The second order ODE:

$(1): \quad y + y = 0$

has the general solution:

$y = C_1 \sin x + C_2 \cos x$

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:

\(\ds \laptrans {y + y}\)

\(=\)

\(\ds \laptrans {y} + \laptrans y\)

Linear Combination of Laplace Transforms

\(\ds \)

\(=\)

\(\ds s^2 \laptrans y - s \map y 0 - \map {y'} 0 + \laptrans y\)

Laplace Transform of Second Derivative

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:

\(\ds y\)

\(=\)

\(\ds \invlaptrans {\map y 0 \frac s {s^2 + 1} + \map {y'} 0 \frac 1 {s^2 + 1} }\)

Definition of Inverse Laplace Transform

\(\ds \)

\(=\)

\(\ds \map y 0 \invlaptrans {\frac s {s^2 + 1} } + \map {y'} 0 \invlaptrans {\frac 1 {s^2 + 1} }\)

Linear Combination of Laplace Transforms

\(\ds \)

\(=\)

\(\ds \map y 0 \invlaptrans {\laptrans {\cos x} } + \map {y'} 0 \invlaptrans {\laptrans {\sin x} }\)

Laplace Transform of Cosine, Laplace Transform of Sine

\(\ds \)

\(=\)

\(\ds \map y 0 \cos x + \map {y'} 0 \sin x\)

Definition of Inverse Laplace Transform

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

$\blacksquare$