Method of Variation of Parameters

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:

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

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.

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

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 $v_2 \left({x}\right)$.