Author Topic: TT1 Problem 3 (main)  (Read 2468 times)

Victor Ivrii

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TT1 Problem 3 (main)
« on: October 16, 2018, 05:25:23 AM »
(a) Find the general solution for equation
\begin{equation*}
y''+5y'+6y=-30e^{2t} + 3 e^{-2t}.
\end{equation*}

(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Jerry Qinghui Yu

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Re: TT1 Problem 3 (main)
« Reply #1 on: October 16, 2018, 08:58:59 AM »
a). First solve the homogenous part
\begin{align*}
y'' + 5y' + 6y &= 0\\
r^2 + 5r + 6 &=0\\
(r+2)(r+3) &=0\\
r_1 = -2,\  r_2 &= -3
\end{align*}
So solution to homogenous part is
\begin{align*}
y_c = c_1e^{-2t} + c_2e^{-3t}
\end{align*}
Next we solve \begin{align}
y'' + 5y' + 6y &= -30e^{2t}
\end{align}
Let
\begin{align*}
Y_1 &= Ae^{2t}, \text{and}\\
Y_1' &= 2Ae^{2t}\\
Y_1'' &= 4Ae^{2t}, \text{Combining with (1)}
\end{align*}
\begin{align*}
4Ae^{2t} + 10Ae^{2t} &+ 6Ae^{2t} = -30e^{2t}\\
A &= -\frac{3}{2}\\
Y_1 &= -\frac{3}{2}e^{2t}
\end{align*}

Next we solve
\begin{align}
y'' + 5y' + 6y &= 3e^{-2t}
\end{align}
Since we already have $e^{-2t}$ in our solution, let
\begin{align*}
Y_2 &= Bte^{-2t}, \text{and}\\
Y_2' &= Be^{-2t} - 2Bte^{-2t}\\
Y_2'' &= -4Be^{-2t} + 4Bte^{-2t}, \text{Combining with (2)}
\end{align*}
\begin{align*}
 (-4Be^{-2t} + 4Bte^{-2t}) + &5(Be^{-2t} - 2Bte^{-2t}) + 6(Bte^{-2t}) = 3e^{-2t}\\
B &= 3\\
Y_2 &=3te^{-2t}
\end{align*}
Combining above we have the general solution
\begin{align}
y &= c_1e^{-2t} + c_2e^{-3t} -\frac{3}{2}e^{2t} + 3te^{-2t}\\
y' &= -2c_1e^{-2t} - 3c_2e^{-3t} - 3e^{2t} +3e^{-2t} - 6te^{-2t}
\end{align}
b). Plug in $y(0)=0, y'(0) = 0$ to (3) and (4)
with
\begin{align*}
c_1 + c_2  &= \frac{3}{2}\\
-2c_1 - 3c_2  &= 0\\
c_1 = \frac{9}{2}&,\ c_2 = -3
\end{align*}
And finally,
\begin{align}
y &= \frac{9}{2}e^{-2t} - 3e^{-3t} -\frac{3}{2}e^{2t} + 3te^{-2t}\\
\end{align}

Zhiya Lou

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Re: TT1 Problem 3 (main)
« Reply #2 on: October 16, 2018, 09:06:33 AM »
scan solution
(sorry, almost post the same time, I didn't notice the perfect solution is posted before me)

Yulin WANG

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Re: TT1 Problem 3 (main)
« Reply #3 on: October 16, 2018, 09:09:43 AM »
In the attachment.

Victor Ivrii

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Re: TT1 Problem 3 (main)
« Reply #4 on: October 18, 2018, 04:13:25 AM »
Jerry did everything right.

Yulin, Zhiya, there was no reason for you to post (even if you typed).