# Linear Second Order ODE/y'' + k^2 y = 0

## Theorem

The second order ODE:

- $(1): \quad y'' + k^2 y = 0$

has the general solution:

- $y = A \, \map \sin {k x + B}$

or can be expressed as:

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

## Proof 1

Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:

- $p \dfrac {\d p} {\d y} = -k^2 y$

where $p = \dfrac {\d y} {\d x}$.

From:

this has the solution:

- $p^2 = -k^2 y^2 + C$

or:

- $p^2 + k^2 y^2 = C$

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:

\(\ds p^2 + k^2 y^2\) | \(=\) | \(\ds k^2 \alpha^2\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds p = \dfrac {\d y} {\d x}\) | \(=\) | \(\ds \pm k \sqrt {\alpha^2 - y^2}\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d y} {\sqrt {\alpha^2 - y^2} }\) | \(=\) | \(\ds \int \pm k \rd x\) | Separation of Variables | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \arcsin \dfrac y \alpha\) | \(=\) | \(\ds \int \pm k x + \beta\) | Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \alpha \map \sin {\pm k x + \beta}\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds A \map \sin {k x + B}\) |

From Multiple of Sine plus Multiple of Cosine: Sine Form, this can be expressed as:

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

$\blacksquare$

## Proof 2

It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.

Its auxiliary equation is:

- $(2): \quad: m^2 + k^2 = 0$

From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:

- $m_1 = k i$
- $m_2 = -k i$

These are complex and unequal.

So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:

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

or, by disposing the constants differently:

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

$\blacksquare$

## Also presented as

This second order ODE can also be presented in the form:

- $\dfrac {\d^2 y} {\d x^2} = -k^2 y$

## Also see

## Sources

- 1952: H.T.H. Piaggio:
*An Elementary Treatise on Differential Equations and their Applications*(revised ed.) ... (previous) ... (next): Chapter $\text I$: Introduction and Definitions. Elimination. Graphical Representation: $1 \ (1)$