Linear Second Order ODE/y'' + y = 0/Proof 3
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
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.15$: The General Solution of the Homogeneous Equation: Example $1$