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Variation of Parameters

For a second-order Ordinary Differential Equation,

y''+p(x)y'+q(x)y = g(x).
\end{displaymath} (1)

Assume that linearly independent solutions $y_1(x)$ and $y_2(x)$ are known. Find $v_1$ and $v_2$ 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
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 $y^*$, ${y^*}'$, and ${y^*}''$ back into the original equation to obtain
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)

v_1'y_1'+v_2'y_2' = g(x).
\end{displaymath} (8)

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

Generalizing to an $n$th degree ODE, let $y_1$, ..., $y_n$ be the solutions to the homogeneous ODE and let $v_1'(x)$, ..., $v_n'(x)$ be chosen such that

y_1v_1'+y_2v_2'+\ldots +y_nv_n' = 0\cr
...{(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
y^*(x) = v_1(x)y_1(x)+\ldots +v_n(x)y_n(x).
\end{displaymath} (12)

© 1996-9 Eric W. Weisstein