Author Topic: Quiz 4 TUT0402  (Read 437 times)

kaye

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Quiz 4 TUT0402
« on: October 18, 2019, 02:00:00 PM »
Question:   Find the general solution of the given differential equation: $y'' - y' - 2y = cosh(2t)$    Hint: $cosh(t) = \frac{e^t + e^{-t}}{2}$



Write the differential equation into:
$$y'' - y' -2y = \frac{1}{2}e^{2t} + \frac{1}{2}e^{-2t}$$

homogeneous solution:
\begin{align}
    r^2 - r - 2 &= 0 \notag\\
    (r - 2)(r + 1) &= 0 \notag
\end{align}

$r_1 = 2$ and $r_2 = -1 \implies y_c = c_1e^{2t} + c_2e^{-t}$



$y'' - y' - 2y = \frac{1}{2}e^{2t} $:
\begin{align}
    y_{p1} &= Ate^{2t}\notag\\
    y' &= (A + 2At)e^{2t}\notag\\
    y'' &= (4A + 4At)e^{2t}\notag
\end{align}
\begin{align}
    (4A + 4At)e^{2t} - (A + 2At)e^{2t} - 2Ate^{2t} &= \frac{1}{2}e^{2t}\notag\\
    4A + 4At - A - 2At - 2At &= \frac{1}{2}\notag\\
    3A &= \frac{1}{2}\notag\\
    A &= \frac{1}{6}\notag
\end{align}

$y'' - y' - 2y = \frac{1}{2}e^{-2t} $:
\begin{align}
    y_{p2} &= Be^{-2t}\notag\\
    y' &= (-2B)e^{2t}\notag\\
    y'' &= (4B)e^{2t}\notag
\end{align}
\begin{align}
    (4B)e^{2t} - (-2B)e^{2t} - 2Be^{-2t} &= \frac{1}{2}e^{-2t}\notag\\
    4B + 2B - 2B &= \frac{1}{2}\notag\\
    4B &= \frac{1}{2}\notag\\
    B &= \frac{1}{8}\notag
\end{align}

from $y_c = c_1e^{2t} + c_2e^{-t}$, $y_{p1} = \frac{1}{6}te^{2t}$, and $y_{p2} = \frac{1}{8}e^{-2t}$ we get:
$$y = c_1e^{2t} + c_2e^{-t} + \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t}$$