### Author Topic: TUT0402 Quiz4  (Read 750 times)

#### Di Qiu

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##### TUT0402 Quiz4
« on: October 18, 2019, 02:00:01 PM »
Find the general solution of y'' - y' - 2y = cosh(2t):
The associated homogenous equation is: \begin{align*} y'' - y' - 2y &= 0 \\ (r-2)(r+1) &= 0 \end{align*}
$$r=2, r=-1$$
Therefore, the complementary function is: $$y_c(t) = c_1e^{-t} + c_2e^{2t}$$
Since $$cosh(t) = \frac{1}{2}(e^t+e^{-t})$$ $$cosh(2t) = \frac{1}{2}(e^{2t}+e^{-2t})$$
we could rewrite the funtion as: $$y'' - y' - 2y = \frac{1}{2}(e^{2t}+e^{-2t})$$
Then we use the method of undetermined coefficients to find the solution for the non-homogeneous function above:
Let $$y_p(t) = Ae^{2t} + Be^{-2t}$$
Then we find $$Ae^{2t}, c_2e^{2t}$$ has the same format, so we times t to get a new equation:
$$y_p(t) = Ate^{2t} + Be^{-2t}$$
Then, $$y_p'(t) = Ae^{2t} + 2Ate^{2t} - 2Be^{-2t}$$ $$y_p''(t) = 2Ae^{2t} + 4Ate^{2t} + 2Ae^{2t} + 4Be^{-2t}$$
Subsititutes back to the equation:
$$2Ae^{2t} + 4Ate^{2t} + 2Ae^{2t} + 4Be^{-2t} -Ae^{2t} - 2Ate^{2t} + 2Be^{-2t} -2Ate^{2t} - 2Be^{-2t} = \frac{1}{2}(e^{2t}+e^{-2t})$$
Then we have: \begin{align*} 2A + 2A - A &= \frac{1}{2} \\ A &= \frac{1}{6} \\ 4B+2B-2B &= \frac{1}{2} \\ B &=\frac{1}{8}\end{align*}
Hence, the particular solution is:
$$y_p(t) = \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t}$$
Finally, \begin{align*} y(t) &= y_c(t) + y_p(t) \\ &= c_1e^{-t} + c_2e^{2t} + \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t} \end{align*}