# 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:

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

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:

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