Author Topic: TT1 Problem 3 (morning)  (Read 2116 times)

Victor Ivrii

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TT1 Problem 3 (morning)
« on: October 16, 2018, 05:29:15 AM »
(a) Find the general solution for equation
\begin{equation*}
y''-5y'+4y=-12e^t + 20 e^{-t}.
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Zhihao Zuo

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Re: TT1 Problem 3 (morning)
« Reply #1 on: October 16, 2018, 09:00:50 AM »
I will type the solution soon!

Zixuan Wang

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Re: TT1 Problem 3 (morning)
« Reply #2 on: October 16, 2018, 09:17:26 AM »
Since the first post did not type out the solution, I will type mine here.
Thank you Shengying, I fixed my problem here.
First, we need to find the homogeneous solution
 $r^2-5r+4=0$
 $(r-4)(r-1)=0$
  $r_1=4,r_2=1$
  $y_c(t) = c_1e^{4t}+c_2e^{t}$
Secondly, we need to find particular solution of $y" - 5y' +4y = -12e^t$
$y_p(t) =Ate^t, y'_p(t) = Ae^t +Ate^t, y"_p(t) = 2Ae^t+Ate^t$
plug in back to the equation we get A=4
Thus, $y_p(t) =4te^t$
Thirdly, we need to find particular solution of $y" - 5y' +4y = 20e^{-t}$
$y_p(t) =Bte^{-t}, y'_p(t) = -Be^{-t}, y"_p(t) = Be^{-t}$
plug in back to the equation we get B=2
Thus, $y_p(t) =2e^{-t}$
Therefore, $y(t) = c_1e^4t +c_2e^t+4te^t+2e^{-t}$
(b) $y(0) = 0$ so that $c_1+c_2 = -2$
     $y'(0) =0$ so that $4c_1+c_2 = -2$
    Therefore $c_1=0, c_2 = -2$ and $y(t) = -2e^t +4te^t+2e^{-t}$
« Last Edit: October 16, 2018, 11:03:43 AM by Zixuan Wang »

Shengying Yang

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Re: TT1 Problem 3 (morning)
« Reply #3 on: October 16, 2018, 10:30:15 AM »
Zixuan, I think you made a mistake.
Plugging in $y(0)=0$ ,  you should get $0=C_1+C_2+2=0$. Therefore, $C_1=1, C_2=-2$
$$∴y=-2e^t+4te^t+2e^{-t}$$

Victor Ivrii

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Re: TT1 Problem 3 (morning)
« Reply #4 on: October 18, 2018, 04:48:01 AM »
Everybody was right:  Zhihao solved correctly,  Zixua typed (you need to learn a better way to type it) and Shengying found a mistype in the solution (not the answer)