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

1. ## 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!

2. let $\displaystyle a(x)y''(x)+b(x)y'(x)+c(x)y(x)=f(x)$ (1) be the non homogeneous second order differential equation.

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

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

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 $\displaystyle c_1'(x)y_1(x)+c_2'(x)y_2(x)=0$ (2); differentiate again to get $\displaystyle y_p''(x)=c'_1(x)y_1'(x)+c_1(x)y''_1(x)+c'_2(x)y_2' (x)+c_2(x)y''_2(x)$ (4), substitute (4) on (1) to get $\displaystyle c_1'(x)y_1'(x)+c_2'(x)y'_2(x)=\dfrac{f(x)}{a(x)}$ (5).

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

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

now we solve it as follows:

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

\displaystyle \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

\displaystyle \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 $\displaystyle y(x_0)=y_0$ and $\displaystyle y'(x_0)=y_1$ and the final expression reads

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