Method of Variation of Parameters

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Proof Technique

The method of variation of parameters is a technique for finding a particular solution to a nonhomogeneous linear second order ODE:

$(1): \quad y'' + P \left({x}\right) y' + Q \left({x}\right) y = R \left({x}\right)$

provided that the general solution of the corresponding homogeneous linear second order ODE:

$(2): \quad y'' + P \left({x}\right) y' + Q \left({x}\right) y = 0$

is already known.


Method

Let the general solution of $(2)$ be:

$y = C_1 y_1 \left({x}\right) + C_1 y_2 \left({x}\right)$


Then a particular solution of $(1)$ is:

$\displaystyle y = y_1 \int -\frac {y_2 \left({x}\right) R \left({x}\right)} {W \left({y_1, y_2}\right)} \, \mathrm d x + y_2 \int \frac {y_1 \left({x}\right) R \left({x}\right)} {W \left({y_1, y_2}\right)} \, \mathrm d x$

where $W \left({y_1, y_2}\right)$ denotes the Wronskian of $y_1 \left({x}\right)$ and $y_2 \left({x}\right)$.


Proof

Let the general solution of $(2)$ be:

$(3): \quad y = C_1 y_1 \left({x}\right) + C_1 y_2 \left({x}\right)$

Let the arbitrary constants $C_1$ and $C_2$ be replaced by functions $v_1 \left({x}\right)$ and $v_2 \left({x}\right)$.

It is required that $v_1$ and $v_2$ be determined so as to make:

$(4): \quad y = v_1 \left({x}\right) y_1 \left({x}\right) + v_2 \left({x}\right) y_2 \left({x}\right)$

a particular solution of $(1)$.

Then:

\(\displaystyle y'\) \(=\) \(\displaystyle \left({v_1 {y_1}' + {v_1}' y_1}\right) + \left({v_2 {y_2}' + {v_2}' y_2}\right)\) Product Rule for Derivatives
\((5):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \left({v_1 {y_1}' + v_2 {y_2}'}\right) + \left({ {v_1}' y_1 + {v_2}' y_2}\right)\)


Suppose ${v_1}' y_1 + {v_2}' y_2$ were made to vanish:

$(6): \quad {v_1}' y_1 + {v_2}' y_2 = 0$


Then:

\((7):\quad\) \(\displaystyle y'\) \(=\) \(\displaystyle v_1 {y_1}' + v_2 {y_2}'\)
\((8):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle y''\) \(=\) \(\displaystyle \left({v_1 {y_1}'' + {v_1}' {y_1}'}\right) + \left({v_2 {y_2}'' + {v_2}' {y_2}'}\right)\) Product Rule for Derivatives


Hence:

\(\displaystyle y'' + P y' + Q y\) \(=\) \(\displaystyle R \left({x}\right)\) $(1):$ given
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left({v_1 {y_1}'' + {v_1}' {y_1}'}\right) + \left({v_2 {y_2}'' + {v_2}' {y_2}'}\right) + P y' + Q y\) \(=\) \(\displaystyle R \left({x}\right)\) substituting from $(8)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left({v_1 {y_1}'' + {v_1}' {y_1}'}\right) + \left({v_2 {y_2}'' + {v_2}' {y_2}'}\right) + P \left({v_1 {y_1}' + v_2 {y_2}'}\right) + Q \left({x}\right) y\) \(=\) \(\displaystyle R \left({x}\right)\) substituting from $(7)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left({v_1 {y_1}'' + {v_1}' {y_1}'}\right) + \left({v_2 {y_2}'' + {v_2}' {y_2}'}\right) + P \left({v_1 {y_1}' + v_2 {y_2}'}\right) + Q \left({v_1 y_1 + v_2 y_2}\right)\) \(=\) \(\displaystyle R \left({x}\right)\) substituting from $(4)$
\((9):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle v_1 \left({ {y_1}'' + P {y_1}' + Q y_1}\right) + v_2 \left({ {y_2}'' + P {y_2}' + Q y_2}\right) + {v_1}' {y_1}' + {v_2}' {y_2}'\) \(=\) \(\displaystyle R \left({x}\right)\) rearranging


Because $y_1$ and $y_2$ are both particular solutions of $(2)$:

${y_1}'' + P {y_1}' + Q y_1 = {y_2}'' + P {y_2}' + Q y_2 = 0$

and so from $(9)$:

$(10): \quad {v_1}' {y_1}' + {v_2}' {y_2}' = R \left({x}\right)$


In summary:

\((6):\quad\) \(\displaystyle {v_1}' y_1 + {v_2}' y_2\) \(=\) \(\displaystyle 0\)
\((10):\quad\) \(\displaystyle {v_1}' {y_1}' + {v_2}' {y_2}'\) \(=\) \(\displaystyle R \left({x}\right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle {v_1}'\) \(=\) \(\displaystyle \frac {y_2 R \left({x}\right)} {y_2 {y_1}' - y_1 {y_2}'}\)
\(\displaystyle {v_2}'\) \(=\) \(\displaystyle \frac {y_1 R \left({x}\right)} {y_1 {y_2}' - y_2 {y_1}'}\)
\((11):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle {v_1}'\) \(=\) \(\displaystyle -\frac {y_2 R \left({x}\right)} {W \left({y_1, y_2}\right)}\)
\(\displaystyle {v_2}'\) \(=\) \(\displaystyle \frac {y_1 R \left({x}\right)} {W \left({y_1, y_2}\right)}\)


We started with the assumption that:

$(3): \quad y = C_1 y_1 \left({x}\right) + C_1 y_2 \left({x}\right)$

and so $y_1$ and $y_2$ are linearly independent.

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

$W \left({y_1, y_2}\right) \ne 0$

and so $(11)$ is defined.


Thus:

\(\displaystyle v_1\) \(=\) \(\displaystyle \int -\frac {y_2 R \left({x}\right)} {W \left({y_1, y_2}\right)} \, \mathrm d x\)
\(\displaystyle v_2\) \(=\) \(\displaystyle \int \frac {y_1 R \left({x}\right)} {W \left({y_1, y_2}\right)} \, \mathrm d x\)


and so as required:

$\displaystyle y = y_1 \int -\frac {y_2 \left({x}\right) R \left({x}\right)} {W \left({y_1, y_2}\right)} \, \mathrm d x + y_2 \int \frac {y_1 \left({x}\right) R \left({x}\right)} {W \left({y_1, y_2}\right)} \, \mathrm d x$

$\blacksquare$


Source of Name

The name method of variation of parameters derives from the method of operation: the parameters $C_1$ and $C_2$ are made to vary by replacing them with the functions $v_1 \left({x}\right)$ and $v_2 \left({x}\right)$.


Sources