Linear Second Order ODE/y'' + y = 0/Proof 1

From ProofWiki
Jump to navigation Jump to search

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:

First Order ODE: $y \rd y = k x \rd x$

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:

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

$\blacksquare$