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

  1. #1
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    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!
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  2. #2
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    Krizalid's Avatar
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    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}<br />
   {{y}_{1}}(x) & {{y}_{2}}(x)  \\<br />
   {{y}_{1}}'(x) & {{y}_{2}}'(x)  \\<br />
\end{matrix} \right)\left( \begin{matrix}<br />
   {{c}_{1}}'(x)  \\<br />
   {{c}_{2}}'(x)  \\<br />
\end{matrix} \right)=\left( \begin{matrix}<br />
   0  \\<br />
   \frac{f(x)}{a(x)}  \\<br />
\end{matrix} \right).

    now we solve it as follows:

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

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

    (2) writes as

    \begin{aligned}<br />
   {{y}_{p}}(x)&=\left( \begin{matrix}<br />
   {{y}_{1}}(x) & {{y}_{2}}(x)  \\<br />
\end{matrix} \right)\left( \begin{matrix}<br />
   {{c}_{1}}(x)  \\<br />
   {{c}_{2}}(x)  \\<br />
\end{matrix} \right) \\ <br />
 & =\int_{{}}^{x}{\frac{f(t)\left( \begin{matrix}<br />
   {{y}_{1}}(x) & {{y}_{2}}(x)  \\<br />
\end{matrix} \right)}{a(t)W\left[ {{y}_{1}},{{y}_{2}} \right](t)}\left( \begin{matrix}<br />
   -{{y}_{2}}(t)  \\<br />
   \phantom{-}{{y}_{1}}(t)  \\<br />
\end{matrix} \right)\,dt} \\ <br />
 & =\int_{{}}^{x}{\frac{f(t)}{a(t)W\left[ {{y}_{1}},{{y}_{2}} \right](t)}\left| \begin{matrix}<br />
   {{y}_{1}}(t) & {{y}_{2}}(t)  \\<br />
   {{y}_{1}}(x) & {{y}_{2}}(x)  \\<br />
\end{matrix} \right|\,dt}. <br />
\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}<br />
   {{y}_{1}}(t) & {{y}_{2}}(t)  \\<br />
   {{y}_{1}}(x) & {{y}_{2}}(x)  \\<br />
\end{matrix} \right|\,dt}.
    Last edited by Krizalid; November 13th 2010 at 02:20 PM.
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