### Author Topic: TT3-LEC0201-ALT-E-Q1  (Read 221 times)

#### RunboZhang

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

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

\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}