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

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


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

this has the general solution:

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


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

As the left hand side is the sum of squares, $C$ has to be positive for this to have any solutions.

Thus, let $C = \alpha^2$.


\(\ds p^2 + k^2 y^2\) \(=\) \(\ds k^2 \alpha^2\)
\(\ds \leadsto \ \ \) \(\ds p = \dfrac {\d y} {\d x}\) \(=\) \(\ds \pm k \sqrt {\alpha^2 - y^2}\)
\(\ds \leadsto \ \ \) \(\ds \int \dfrac {\d y} {\sqrt {\alpha^2 - y^2} }\) \(=\) \(\ds \int \pm k \rd x\) Solution to Separable Differential Equation
\(\ds \leadsto \ \ \) \(\ds \arcsin \dfrac y \alpha\) \(=\) \(\ds \int \pm k x + \beta\) 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:

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


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:

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


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