MAT244--2019F > Term Test 2

Problem 2 (main sitting)

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

a)

Since the coefficient of y'' is 4,

Then, Wronskain is

$w=ce^{-\int4dt}=ce^{-4t}$

b)

Use homogeneous equation to find fundamental solutions,

$y'''+4y''+y'-6y=24e^{t}$

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

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

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

$w = \begin{array}{ccc} e^{t} & e^{-2t} & e^{-3t}\\ e^{t} & -2e^{-2t} & -3e^{-3t}\\ e^{t} & 4e^{-2t} & 9e^{-3t} \end{array}=-12e^{-4t}$

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

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

c)

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

$y'''+4y''+y'-6y=24e^{t}$

$y_{p}(t)=Ate^{t}$

$y'_{p}(t)=Ae^{t}+Ate^{t}$

$y''_{p}(t)=2Ae^{t}+Ate^{t}$

$y'''_{p}(t)=3Ae^{t}+Ate^{t}$

Plug into the equation,

$3Ae^{t}+Ate^{t}+4(2Ae^{t}+Ate^{t})+Ae^{t}+Ate^{t}-6Ate^{t}=12Ae^{t}=24e^{t}$

We get A=2

$y_{p}(t)=2te^{t}$

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

nayan:
$\text(a)\\$

$\text{Using abels theorem,} \\ W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}\\$
$\text(b)\\$
$\text{We have the following characteristic polynomial,}\\$
$r^3+4r^2+r-6=0\\$
$(r-1)(r^2+5r+6)=0\\$
$r=1, r=-2, r=-3\\$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}\\$
$W=\begin{vmatrix} e^t & e^{-2t} & e^{-3t}\\ e^t & -2e^{-2t} & -3e^{-3t}\\ e^t & 4e^{-2t} & 9e^{-3t}\\ \end{vmatrix}=-12e^{-4t}\\$
$\text{$\therefore c=-12$}\\$
$\text(c)\\$
$y_p(t)=Ate^t\\$
$y_p^{\prime}(t)=Ae^t+Ate^t\\$
$y_p^{\prime\prime}(t)=2Ae^t+Ate^t\\$
$y_p^{\prime\prime\prime}(t)=3Ae^t+Ate^t\\$
$3Ae^t+Ate^t+8Ae^t+4Ate^t+Ae^t+Ate^t-6Ate^t=12Ae^t\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad 12A=24\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad A=2$
$\text{We finally have the general solution,}\\ y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$

Mingdi Xie:
This is my solution :)