# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 20, 2018, 05:45:04 AM

Title: TT2-P2
Post by: Victor Ivrii on November 20, 2018, 05:45:04 AM
Consider equation

y'''+y''  -y'-y= 8e^{-t}.
\label{2-1}

(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}).
Title: Re: TT2-P2
Post by: Tzu-Ching Yen on November 20, 2018, 07:02:49 AM
a)
$\frac{dW}{W} = -1 dt$
$W = e^{\int -1 dt} = ce^{-t}$

$r^3 + r^2 - r - 1 = (r^2 - 1)(r+1) = (r+1)^2(r-1)$
$r = 1, -1$ 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
$y = c_1e^{-t} + c_2te^{-t} + c_3e^t - 2t^2e^{-t}$
Title: Re: TT2-P2
Post by: Blair Zhang 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.
Title: Re: TT2-P2
Post by: Boyu Zheng 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*}
$$6Ae^{-t}t-6Ae^{-t}-Ae^{-t}t^2-4Ae^{-t}+2Ae^{-t}+Ae^{-t}t^2-2Ate^{-t}+At^2e^{-t}-At^2e^{-t}=8e^{-t}$$
$$-4A=8$$
$$A=-2$$
$$y(t)=c_1e^t+c_2e^{-t}+c_3e^{-t}t-2e^{-t}$$
Title: Re: TT2-P2
Post by: Tzu-Ching Yen 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.