Author Topic: TUt0402 question  (Read 460 times)

shangluy

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TUt0402 question
« on: October 18, 2019, 02:00:36 PM »
Solve $y'' - y' - 2y = cosh2t$

Firstly, consider homogeneous equation $y'' - y' - 2y = 0$, the characteristic polynomial is
\begin{align*}
r^2 - r - 2 &= 0\\
(r + 1)(r - 2) &= 0
\end{align*}

Therefore, the solution to homogeneous equation is $c_1 e^{-t} + c_2 e^{2t}$

Then need to solve $y'' - y' - 2y = cosh2t$, since
\begin{equation*}
    cosh(2t) = \frac{e^{2t} + e^{-2t}}{2}
\end{equation*}

we come to guess $y = Ate^{2t} + Be^{-2t}$ is a solution.
\begin{align*}
    y' &= Ae^{2t} + 2Ate^2t - 2Be^{-2t}\\
    y'' &= 2Ae^{2t} + 2Ae^{2t} + 4Ate^{2t} + 4Be^{-2t}\\
    &= (4A + 4At)e^{2t} + 4Be^{-2t}
\end{align*}
then plug $y$, $y'$ and $y''$ back in $y'' - y' - 2y$, we get

\begin{align*}
    y'' - y' - 2y =& (4A + 4At)e^{2t} + 4Be^{-2t} - (Ae^{2t} + 2Ate^2t - 2Be^{-2t}) - 2(Ate^{2t} + Be^{-2t}) \\
    &= (4A - A)e^{2t} + (4A - 2A - 2A)te^{2t} + (4B + 2B - 2B)e^{-2t}\\\
    &= 3Ae^{2t} + 4Be^{-2t}
\end{align*}
since $y'' - y' - 2y = cosh2t$
\begin{align*}
    &3Ae^{2t} + 4Be^{-2t} = \frac{e^{2t} + e^{-2t}}{2}\\
    &\implies 3A = \frac{1}{2}\quad and\quad 4B = \frac{1}{2}\\
    &\implies A = \frac{1}{6}\quad and \quad B = \frac{1}{8}
\end{align*}
Therefore the solution to $y'' - y' - 2y = cosh2t$ is $y = \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t}$
Then, the general solution is
\begin{equation*}
    y = c_1 e^{-t} + c_2 e^{2t} + \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t}
\end{equation*}