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

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE:

$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$

on a closed interval $\closedint a b$.


Let $y_1$ and $y_2$ be linearly independent.


Then the general solution to $(1)$ is:

$y = C_1 \map {y_1} x + C_2 \map {y_2} x$

where $C_1 \in \R$ and $C_2 \in \R$ are arbitrary constants.


Proof

Let $\map y x$ be any particular solution to $(1)$ on $\closedint a b$.

It is to be shown that constants $C_1$ and $C_2$ can be found such that:

$\map y x = C_1 \map {y_1} x + C_2 \map {y_2} x$

for all $x \in \closedint a b$.

By Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions:

a particular solution to $(1)$ over $\closedint a b$ is completely determined by:
its value
and:
the value of its derivative

at a single point.

From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE:

$C_1 \map {y_1} x + C_2 \map {y_2} x$

is a particular solution to $(1)$ over $\closedint a b$

We also have:

$\map y x$

is a particular solution to $(1)$ over $\closedint a b$

Thus it is sufficient to prove that:

$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that:
$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$
and:
$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$


For this system to be solvable for $C_1$ and $C_2$ it is necessary that:

$\begin{vmatrix} \map {y_1} x & \map {y_2} x \\ \map { {y_1}'} x & \map { {y_2}'} x \\ \end{vmatrix} = \map {y_1} x \map { {y_2}'} x - \map {y_2} x \map { {y_1}'} x \ne 0$

That is, that the Wronskian $\map W {y_1, y_2} \ne 0$ at $x_0$.


From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE:

if $\map W {y_1, y_2} \ne 0$ at $x_0$, then $\map W {y_1, y_2} \ne 0$ for all $x \in \closedint a b$.

Hence it does not matter what point is taken for $x_0$; if the Wronskian is non-zero at one such point, it will be non-zero for all such points.


From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:

$W \left({y_1, y_2}\right) = 0$ for all $x \in \closedint a b$ if and only if $y_1$ and $y_2$ are linearly dependent.

But we have that $y_1$ and $y_2$ are linearly independent.

Hence:

$\forall x \in \closedint a b: \map W {y_1, y_2} \ne 0$

and so:

$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that:
$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$
and:
$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$

The result follows.

$\blacksquare$


Sources