Linear Second Order ODE/y'' + y = 0/Proof 1
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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
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
- $p \dfrac {\d p} {\d y} = -y$
where $p = \dfrac {\d y} {\d x}$.
From:
with $k = 1$, this has the solution:
- $p^2 = -y^2 + C$
or:
- $p^2 + y^2 = C$
As the left hand side is the sum of squares, $C$ has to be positive for this to have any solutions.
Thus, let $C = \alpha^2$.
Then:
\(\ds p^2 + y^2\) | \(=\) | \(\ds \alpha^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds p = \dfrac {\d y} {\d x}\) | \(=\) | \(\ds \pm \sqrt {\alpha^2 - y^2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d y} {\sqrt {\alpha^2 - y^2} }\) | \(=\) | \(\ds \int \pm 1 \rd x\) | Solution to Separable Differential Equation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \arcsin \dfrac y \alpha\) | \(=\) | \(\ds \pm x + \beta\) | Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \alpha \map \sin {\pm x + \beta}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds A \map \sin {x + B}\) |
From Multiple of Sine plus Multiple of Cosine: Sine Form, this can be expressed as:
- $y = C_1 \sin x + C_2 \cos x$
$\blacksquare$