Linear Second Order ODE/y'' + y = 0/Proof 5
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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.
$\blacksquare$