MAT244--2019F > Term Test 2

Problem 2 (main sitting)

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Ruodan Chen:
2) $y'''+4y''+y'-6y=24e^{t}$


Since the coefficient of y'' is 4,

Then, Wronskain is



Use homogeneous equation to find fundamental solutions,


Then $r^{3}+4r^{2}+r-6=0$

Then $r=1, r=-2, r=-3$

Then the solution is $y_{c}(t)=c_{1}e^{t}+c_{2}e^{-2t}+c_{3}e^{-3t}$

$w = \begin{array}{ccc}
e^{t} & e^{-2t} & e^{-3t}\\
e^{t} & -2e^{-2t} & -3e^{-3t}\\
e^{t} & 4e^{-2t} & 9e^{-3t}

$w = -12e^{-4t} =ce^{-4t}$

This is consistent with what we get in part (a) with $c=-12$


Use under determined coefficients method to find the general equation for y(t)






Plug into the equation,


We get A=2


Thus, $y(t)=c_{1}e^{t}+c_{2}e^{-2t}+c_{3}e^{-3t}+2te^{t}$


$\text{Using abels theorem,}
W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}\\$
$\text{We have the following characteristic polynomial,}\\$
$r=1, r=-2, r=-3\\$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}\\$
e^t & e^{-2t} & e^{-3t}\\
e^t &  -2e^{-2t} & -3e^{-3t}\\
e^t & 4e^{-2t} & 9e^{-3t}\\
$\text{$\therefore c=-12$ }\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad 12A=24\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad A=2$
$\text{We finally have the general solution,}\\ y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$

Mingdi Xie:
This is my solution :)


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