Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another
Theorem
Let $\map {y_1} x$ be a particular solution to the homogeneous linear second order ODE:
- $(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
such that $y_1$ is not the trivial solution.
Then there exists a standard procedure to determine another particular solution $\map {y_2} x$ of $(1)$ such that $y_1$ and $y_2$ are linearly independent.
Proof
Let $\map {y_1} x$ be a non-trivial particular solution to $(1)$.
Thus, for all $C \in \R$, $C y_1$ is also a non-trivial particular solution to $(1)$.
Let $\map v x$ be a function of $x$ such that:
- $(2): \quad \map {y_2} x = \map v x \, \map {y_1} x$
is a particular solution to $(1)$ such that $y_1$ and $y_2$ are linearly independent.
Thus:
- $(3): \quad {y_2}'' + \map P x {y_2}' + \map Q x y = 0$
Differentiating $(2)$ twice with respect to $x$:
- ${y_2}' = \map {v'} x \, \map {y_1} x + \map v x \, \map { {y_1}'} x$
and:
- ${y_2}'' = \map {v''} x \, \map {y_1} x + 2 \, \map {v'} x \, \map { {y_1}'} x + \map v x \, \map { {y_1}''} x$
Substituting these into $(3)$ and rearranging:
- $v paren { {y_1}'' + P {y_1}' + Q y_1} + v'' y_1 + v' \paren {2 {y_1}'' + P y_1} = 0$
Since $y_1$ is a particular solution of $(1)$, this reduces to:
- $v'' y_1 + v' \paren {2 {y_1}'' + P y_1} = 0$
which can be expressed as:
- $\dfrac {v''} {v'} = -2 \dfrac { {y_1}'} {y_1} - P$
Integration gives:
- $\ds \ln v' = - 2 \ln y_1 - \int P \rd x$
This leads to:
- $v' = \dfrac 1 { {y_1}^2} e^{-\int P \rd x}$
and so:
- $\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
From Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution, $y_1$ and $v y_1$ are linearly independent.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: The Use of a Known Solution to find Another