method of variation of parameters to find particular solution of this simple diff eqn

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• Oct 12th 2010, 11:12 PM
qtpipi
method of variation of parameters to find particular solution of this simple diff eqn
hi guys,

haven't done diff eqn stuff in a long time. i am not sure how to do this:

find a particular solution to this differential equation by using the method of variation of parameters:

y'' - 4y = sinh(2x)

i know the answer is y_p(x) = (1/16)(4*x*cosh(2x) - sinh(2x)) but can someone show me how to reach this solution??!!

thanks a lot!
• Oct 13th 2010, 04:18 PM
Krizalid
let $a(x)y''(x)+b(x)y'(x)+c(x)y(x)=f(x)$ (1) be the non homogeneous second order differential equation.

we have that $y_p(x)=c_1(x)y_1(x)+c_2(x)y_2(x)$ (2) whereat $y_1,y_2$ are the solutions of the homogeneous equation.

we differentiate the equation (3) and we get $y'_p(x)=c_1'(x)y_1(x)+c_2'(x)y_2(x)+c_1(x)y_1'(x)+ c_2(x)y'_2(x).$

by doing this, we would get two particular solutions for (1), but we actually want just one solution, so to get these thing work okay, we put $c_1'(x)y_1(x)+c_2'(x)y_2(x)=0$ (2); differentiate again to get $y_p''(x)=c'_1(x)y_1'(x)+c_1(x)y''_1(x)+c'_2(x)y_2' (x)+c_2(x)y''_2(x)$ (4), substitute (4) on (1) to get $c_1'(x)y_1'(x)+c_2'(x)y'_2(x)=\dfrac{f(x)}{a(x)}$ (5).

now (2) and (5) form the following system of equations:

$\left( \begin{matrix}
{{y}_{1}}(x) & {{y}_{2}}(x) \\
{{y}_{1}}'(x) & {{y}_{2}}'(x) \\
\end{matrix} \right)\left( \begin{matrix}
{{c}_{1}}'(x) \\
{{c}_{2}}'(x) \\
\end{matrix} \right)=\left( \begin{matrix}
0 \\
\frac{f(x)}{a(x)} \\
\end{matrix} \right).$

now we solve it as follows:

$\left( \begin{matrix}
{{c}_{1}}'(x) \\
{{c}_{2}}'(x) \\
\end{matrix} \right)=\dfrac{1}{W\left[ {{y}_{1}},{{y}_{2}} \right](x)}\left( \begin{matrix}
\phantom{-}{{y}_{2}}'(x) & -{{y}_{2}}(x) \\
-{{y}_{1}}'(x) & \phantom{-}{{y}_{1}}(x) \\
\end{matrix} \right)\left( \begin{matrix}
0 \\
\frac{f(x)}{a(x)} \\
\end{matrix} \right),$

integrate to get

\begin{aligned}
\left( \begin{matrix}
{{c}_{1}}(x) \\
{{c}_{2}}(x) \\
\end{matrix} \right)&=\int_{{}}^{x}{\frac{1}{W\left[ {{y}_{1}},{{y}_{2}} \right](t)}\left( \begin{matrix}
\phantom{-}{{y}_{2}}'(t) & -{{y}_{2}}(t) \\
-{{y}_{1}}'(t) & \phantom{-}{{y}_{1}}(t) \\
\end{matrix} \right)\left( \begin{matrix}
0 \\
\frac{f(t)}{a(t)} \\
\end{matrix} \right)\,dt} \\
& =\int_{{}}^{x}{\frac{f(t)}{a(t)W\left[ {{y}_{1}},{{y}_{2}} \right](t)}\left( \begin{matrix}
-{{y}_{2}}(t) \\
\phantom{-}{{y}_{1}}(t) \\
\end{matrix} \right)\,dt}.
\end{aligned}

(2) writes as

\begin{aligned}
{{y}_{p}}(x)&=\left( \begin{matrix}
{{y}_{1}}(x) & {{y}_{2}}(x) \\
\end{matrix} \right)\left( \begin{matrix}
{{c}_{1}}(x) \\
{{c}_{2}}(x) \\
\end{matrix} \right) \\
& =\int_{{}}^{x}{\frac{f(t)\left( \begin{matrix}
{{y}_{1}}(x) & {{y}_{2}}(x) \\
\end{matrix} \right)}{a(t)W\left[ {{y}_{1}},{{y}_{2}} \right](t)}\left( \begin{matrix}
-{{y}_{2}}(t) \\
\phantom{-}{{y}_{1}}(t) \\
\end{matrix} \right)\,dt} \\
& =\int_{{}}^{x}{\frac{f(t)}{a(t)W\left[ {{y}_{1}},{{y}_{2}} \right](t)}\left| \begin{matrix}
{{y}_{1}}(t) & {{y}_{2}}(t) \\
{{y}_{1}}(x) & {{y}_{2}}(x) \\
\end{matrix} \right|\,dt}.
\end{align}

add initial conditions $y(x_0)=y_0$ and $y'(x_0)=y_1$ and the final expression reads

${{y}_{p}}(x)=\displaystyle\int_{{{x}_{0}}}^{x}{\fr ac{f(t)}{a(t)W\left[ {{y}_{1}},{{y}_{2}} \right](t)}\left| \begin{matrix}
{{y}_{1}}(t) & {{y}_{2}}(t) \\
{{y}_{1}}(x) & {{y}_{2}}(x) \\
\end{matrix} \right|\,dt}.$