Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another

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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