Author Topic: TT2-P2  (Read 996 times)

Victor Ivrii

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TT2-P2
« on: November 20, 2018, 05:45:04 AM »
Consider equation
\begin{equation}
y'''+y''  -y'-y= 8e^{-t}.
\label{2-1}
\end{equation}
(a)  Write a differential equation for the Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

(b)  Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).

(c) Find the general solution of (\ref{2-1}).
« Last Edit: November 20, 2018, 05:49:10 AM by Victor Ivrii »

Tzu-Ching Yen

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Re: TT2-P2
« Reply #1 on: November 20, 2018, 07:02:49 AM »
a)
$\frac{dW}{W} = -1 dt$
$W = e^{\int -1 dt} = ce^{-t}$

b) Characteristic equation reads
$r^3 + r^2 - r - 1 = (r^2 - 1)(r+1) = (r+1)^2(r-1)$
$r = 1, -1$ where $-1$ is a repeated eigenvalue, hence the solutions are
$y_1 = e^{-t}, y_2 = te^{-t}, y_3 = e^t$
After some row operations,
$ W =
e^{-t}det \bigl(\left[ {\begin{array}{ccc}
    1 & t & 1 \\
    0 & 1 & 2 \\
    0 & -2 & 0 \\
\end{array} } \right]\bigr) = 4e^{-t}
$
This agree with part a) where $c = 4$

c) Since $e^{-t}, te^{-t}$ are solutions to homogeneous equation, the form of particular solution is $At^2e^{-t}$, where
$L''(-1) = -4$
$AL''(-1) = 8, A = -2$
Hence the solution is
$y = c_1e^{-t} + c_2te^{-t} + c_3e^t - 2t^2e^{-t}$
« Last Edit: November 20, 2018, 07:42:58 AM by Thomson Yen »

Blair Zhang

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Re: TT2-P2
« Reply #2 on: November 20, 2018, 07:24:29 AM »
I also got c=4 for part b. But I am curious is there any relation between part b and c. My first attempt was using variation of parameter to solve instead of set a particular solution.

Boyu Zheng

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Re: TT2-P2
« Reply #3 on: November 20, 2018, 07:38:42 AM »
$$
a)w=ce^{-\int p(t)   \mathrm{d}t}=ce^{-\int 1\mathrm{d}t}=ce^{-t}
$$
$$
b)\gamma^3+\gamma^2-\gamma-1=0
$$
\begin{align*}
(\gamma-1)(\gamma^2+2\gamma+1)=0\\
\gamma_1=1,\gamma_2=\gamma_3=-1\\
yc(t)=c_1e^t+c_2e^{-t}+c_3e^{-t}t
\end{align*}
\begin{align*}
w&=      \begin{vmatrix}
   e^t   &      e^{-t}       &te^{-t}\\
   e^t   &      -e^{-t}       &e^{-t}-te^{-t}\\
   e^t   &      e^{-t}       &-2e^{-t}+te^{-t}
      \end{vmatrix}   
=e^t(-e^{-t}(-2e^{-t}+te^{-t})-e^{-t}(e^{-t}-te^{-t}))\\
&-e^t(e^{-t}(-2e^{-t}+te^{-t})-e^{-t}(te^{-t}))+e^t(e^{-t}(e^{-t}-te^{-t})+e^{-t}(te^{-t}))\\
&=4e^{-t}
\end{align*}
\centerline{Compared with(a),c=4}
\begin{align*}
c) yp(t)&=At^2e^{-t}\\
y'p(t)&=2Ate^{-t}-At^2e^{-t}\\
y''p(t)&=A(-4e^{-t}t+2e^{-t}+e^{-t}t^2)\\
y'''p(t)&=A(6e^{-t}t-6e^{-t}-e^{-t}t^2)\\
y'''+y''&-y'-y=8e^{-t}
\end{align*}
$$
6Ae^{-t}t-6Ae^{-t}-Ae^{-t}t^2-4Ae^{-t}+2Ae^{-t}+Ae^{-t}t^2-2Ate^{-t}+At^2e^{-t}-At^2e^{-t}=8e^{-t}
$$
$$
-4A=8
$$
$$
A=-2
$$
$$
y(t)=c_1e^t+c_2e^{-t}+c_3e^{-t}t-2e^{-t}
$$

Tzu-Ching Yen

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Re: TT2-P2
« Reply #4 on: November 20, 2018, 07:45:15 AM »
Blair, I don't think there should be any relationship between b) and c). Since Wronskian is only dependent on solutions to homogeneous equation while particular solution is dependent on g(t) ($8e^t$ in this case). Not sure thou.