Linear Second Order ODE/y'' + y = 0/Proof 3

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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

We have that:

\(\ds \frac \d {\d x} \, \sin x\) \(=\) \(\ds \cos x\) Derivative of Sine Function
\(\ds \frac \d {\d x} \, \cos x\) \(=\) \(\ds -\sin x\) Derivative of Cosine Function
\(\ds \leadsto \ \ \) \(\ds \frac {\d^2} {\d x^2} \, \cos x\) \(=\) \(\ds -\cos x\)
\(\ds \frac {\d^2} {\d x^2} \, \sin x\) \(=\) \(\ds -\sin x\)


Hence it can be seen by inspection that:

\(\ds \map {y_1} x\) \(=\) \(\ds \sin x\)
\(\ds \map {y_2} x\) \(=\) \(\ds \cos x\)

are particular solutions to $(1)$.

It is noted that:

$\dfrac {y_1} {y_2} = \tan x$

which is not a constant function on any closed interval $\closedint a b$.


Calculating the Wronskian of $y_1$ and $y_2$:

\(\ds \map W {y_1, y_2}\) \(=\) \(\ds \begin{vmatrix} \sin x & \cos x \\ \cos x & -\sin x \end{vmatrix}\)
\(\ds \) \(=\) \(\ds -\sin^2 x - \cos^2 x\)
\(\ds \) \(=\) \(\ds -1\) Sum of Squares of Sine and Cosine

So the Wronskian of $y_1$ and $y_2$ is never zero.


We have that $(1)$ is a homogeneous linear second order ODE in the form:

$y'' + \map P x y' + \map Q x y = 0$

where $\map P x = 0$ and $\map Q x$ are constant functions.

So by Constant Real Function is Continuous: $P$ and $Q$ are continuous on $\closedint a b$

Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:

$(1)$ has the general solution:
$y = C_1 \sin x + C_2 \cos x$
on $\closedint a b$.

Since $\closedint a b$ can be extended indefinitely without introducing points where $P$ or $Q$ are discontinuous, the general solution is valid for all $x$.

$\blacksquare$


Sources