Author Topic: TT3-LEC0201-ALT-E-Q1  (Read 207 times)

RunboZhang

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TT3-LEC0201-ALT-E-Q1
« on: November 19, 2020, 12:45:28 PM »
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\textbf{Problem 1: } \\\\
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$\text{Consider the equation}$
$$ y''' - 9y'' + 27y' -27y = 24e^{3t}$$
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\textbf{a:}
\text { Write a differential equation for Wronskian of } y_1 ,\ y_2 ,\ y_3 ,\ \text{ which are solutions for homogeneous equation and solve it.}\\\\ \\\\
\textbf{b:}
\text { Find fundamental system } {y_1, y_2, y_3} \text{of solutions for homogeneous equation, and find their Wronskian. Compare with part (a).} \\\\ \\\\
\textbf{c:}
\text { Find the general solution of the equation.} \\\\
$


$
\textbf{Solution for part (a):} \\\\
$

$
\begin{gather}
\begin{aligned}
W &= ce^{- \int{-9} \,dt} \\\\
&= ce^{\int{9}\, dt} \\\\
&= ce^{9t}
\end{aligned}
\end{gather}
$

$
\textbf{Solution for part (b):} \\\\
$

$\text{We have characteristic polynomial: }$
$$\lambda ^{3} - 9\lambda ^{2} + 27 \lambda -27 = 0$$
$\text{Then solve it we have: }$ $$ \lambda_1 =\lambda_2 =\lambda_3 = 3$$
$\text{Hence: }$
$$y_1 = c_1e^{3t} \\\\
y_2 = c_2 t e^{3t} \\\\
y_3 = c_3 t^{2}e^{3t}$$

$\text{Take } c_1 =c_2 =c_3 = 1 \text{, we have:}$

$
\begin{gather}
\begin{aligned}

w_1 &= \begin{bmatrix}
te^{3t} & t^{2}e^{3t}  \\
(3t+1)e^{3t} & (3t^{2}+2t)e^{3t} \\
\end{bmatrix} \\\\
&= (3t^{3}+2t^{2})e^{6t} - (3t^{3}+t^{2})e^{6t} \\\\
&= t^{2}e^{6t}

\end{aligned}
\end{gather}
$

$
\begin{gather}
\begin{aligned}

w_2 &= - \begin{bmatrix}
 e^{3t}& t^{2}e^{3t}  \\
3e^{3t} & (3t^{2}+2t)e^{3t} \\
\end{bmatrix} \\\\
&= -[(3t^{2} + 2t)e^{6t} -3t^{2}e^{6t}] \\\\
&= -2te^{6t}

\end{aligned}
\end{gather}
$

$
\begin{gather}
\begin{aligned}

w_3 &= \begin{bmatrix}
 e^{3t}& te^{3t}  \\
3e^{3t} & (3t+1)e^{3t} \\
\end{bmatrix} \\\\
&= (3t+1)e^{6t}-3te{6t} \\\\
&= e^{6t}

\end{aligned}
\end{gather}
$

$\text{Therefore we have: }$
$
\begin{gather}
\begin{aligned}

W &= y_1''w_1 +y_2''w_2 + y_3''w_3 \\\\
&= (e^{3t})'' t^{2}e^{6t} + (te^{3t})''(-2te^{6t}) + (t^{2}e^{3t})''e^{6t} \\\\
&= 9t^{2}e^{9t}-18t^{2}e^{9t} -12te^{9t} +9t^{2}e^{9t} +12te^{9t} + 2e{9t} \\\\
&= 2e^{9t}

\end{aligned}
\end{gather}
$
$\text{Compared to part(a), Wronskian in part (b) satisfies that of in part (a) and c = 2.}$


$
\textbf{Solution for part (c):} \\\\
\text{using the method of variation, we have}
$
$
\begin{gather}
\begin{aligned}

u_1 & = \int{\frac{24e^{3t}\cdot t^{2}e^{6t}}{2e^{9t}}} \,dt \\\\
&= 12 \int{\frac{ t^{2}e^{9t}}{e^{9t}}} \,dt \\\\
&= 4t^{3} + c_1

\end{aligned}
\end{gather}
$

$
\begin{gather}
\begin{aligned}

u_2 & = \int{\frac{24e^{3t}\cdot (-2)te^{6t}}{2e^{9t}}} \,dt \\\\
&= -24 \int{t} \,dt \\\\
&= -12t^{2} + c_2

\end{aligned}
\end{gather}
$

$
\begin{gather}
\begin{aligned}

u_3 & =  \int{\frac{24e^{3t}\cdot e^{6t}}{2e^{9t}}} \,dt \\\\
& = 12 \int{1} \,dt \\\\
&= 12t + c_3

\end{aligned}
\end{gather}
$

$\text{Therefore the general solution is: }$

$
\begin{gather}
\begin{aligned}
y &= u_1y_1 + u_2y_2 + u_3y_3 \\\\
&= (4t^{3} + c_1)e^{3t} + (-12t^{2} + c_2)te^{3t} + (12t + c_3)t^{2}e^{3t} \\\\
&= c_1e^{3t} + c_2te^{3t} +c_3t^{2}e^{3t}+4t^{3}e^{3t}
\end{aligned}
\end{gather}
$