Author Topic: TT1 Problem 4 (noon)  (Read 1745 times)

Victor Ivrii

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TT1 Problem 4 (noon)
« on: October 16, 2018, 05:33:08 AM »
Find the general solution for equation
\begin{equation*}
y''(t)+6y'(t)+10y(t)=5e^{-3t} +13\cos(t) .
\end{equation*}

Jialu Lin

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Re: TT1 Problem 4 (noon)
« Reply #1 on: October 16, 2018, 06:51:31 AM »
Here is my solution.
« Last Edit: October 16, 2018, 07:00:28 AM by Jialu Lin »

Shengying Yang

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Re: TT1 Problem 4 (noon)
« Reply #2 on: October 16, 2018, 07:37:25 AM »
For the homogeneous part, I think $r=-3\pm i​$ . Therefore, $$y_c(t)=C_1e^{-3t}cost+C_2e^{-3t}sint$$
I will post my answer below.

Shengying Yang

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Re: TT1 Problem 4 (noon)
« Reply #3 on: October 16, 2018, 07:41:48 AM »
First, we consider $y''+6y'+10y=0​$
$$r^2+6r+10=0$$
$$∴r=-3\pm i​$$
$$∴y_c(t)=C_1e^{-3t}\cos t+C_2e^{-3t}\sin t$$
Second, we consider $y''+6y'+10y=5e^{-3t}​$, let $$y_{p1}(t)= Ae^{-3t}, y'_{p1}(t)=-3Ae^{-3t}, y''_{p1}(t)=9Ae^{-t}​$$
$$9Ae^{-3t} -18Ae^{-3t}+10Ae^{-3t} =5e^{-3t}$$
$$∴A=5$$
$$∴y_{p1}(t)= 5e^{-3t}$$
Third, we consider $y''+6y'+10y=13\cos t$ , let $$y_{p2}(t)= B\cos t+C\sin t, y'_{p2}(t)= -B\sin t+C\cos t,y''_{p2}(t)= -B\cos t-C\sin t​$$
$$∴-B\cos t-C\sin t+(-B\sin t+C\cos t)+10(B\cos t+C\sin t)=13\cos t$$
$$∴9C-6B=0, 9B+6C=13 $$
$$∴ B=1, C=\frac{2}{3}$$
$$∴y_{p2}(t)= \cos t+\frac{2}{3}\sin t​$$
Therefore,
$$y(t)=y_c(t)+y_{p1}(t)+y_{p2}(t) =C_1e^{-3t}\cos t+C_2e^{-3t}\sin t+5e^{-3t}+\cos t+\frac{2}{3}\sin t$$

Victor Ivrii

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Re: TT1 Problem 4 (noon)
« Reply #4 on: October 18, 2018, 03:46:10 AM »
Jialu found a correct particular solution for inhomogeneous equation, but made a mistake in the general solution for homogeneous equation. Shengying did everything right (and the LaTeX is good).