Variation of Parameters

For a second-order Ordinary Differential Equation,

$\begin{displaymath} y''+p(x)y'+q(x)y = g(x). \end{displaymath}$

(1)

Assume that linearly independent solutions

and

are known. Find

and

such that

$\displaystyle {y^*}(x)$	$\textstyle =$	$\displaystyle v_1(x)y_1(x)+v_2(x)y_2(x)$	(2)
$\displaystyle {y^*}'(x)$	$\textstyle =$	$\displaystyle (v_1'+v_2'y_2)+(v_1y_1'+v_2y_2').$	(3)

Now, impose the additional condition that

$\begin{displaymath} v_1'y_1+v_2'y_2 = 0 \end{displaymath}$

(4)

so that

$\displaystyle {y^*}'(x)$	$\textstyle =$	$\displaystyle (v_1y_1'+v_2y_2')$	(5)
$\displaystyle {y^*}''(x)$	$\textstyle =$	$\displaystyle v_1'y_1'+v_2'y_2'+v_1y_1''+v_2y_2'.$	(6)

Plug

, ${y^*}'$ , and ${y^*}''$ back into the original equation to obtain

$\begin{displaymath} v_1(y_1''+py_1'+qy_1)+v_2(y_2''+py_2'+qy_2)+v_1'y_1'+v_2'y_2' = g(x) \end{displaymath}$

(7)

$\begin{displaymath} v_1'y_1'+v_2'y_2' = g(x). \end{displaymath}$

(8)

Therefore,

$\displaystyle v_1'y_1+v_2'y_2$	$\textstyle =$	$\displaystyle 0$	(9)
$\displaystyle v_1'y_1'+v_2'y_2'$	$\textstyle =$	$\displaystyle g(x).$	(10)

Generalizing to an th degree ODE, let , ..., be the solutions to the homogeneous ODE and let , ..., be chosen such that

$\begin{displaymath} \cases{ y_1v_1'+y_2v_2'+\ldots +y_nv_n' = 0\cr y_1'v_1'+y_... ...{(n-1)}v_1'+y_2^{(n-1)}v_2'+\ldots+y_n^{(n-1)}v_n' = g(x).\cr} \end{displaymath}$

(11)

Then the particular solution is

$\begin{displaymath} y^*(x) = v_1(x)y_1(x)+\ldots +v_n(x)y_n(x). \end{displaymath}$

(12)