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Term Test 2 / Re: Problem 2 (main sitting)
« on: November 19, 2019, 01:59:47 PM »
$\text(a)\\$
$\text{Using abels theorem,}
\\
W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}\\$
$\text(b)\\$
$\text{We have the following characteristic polynomial,}\\$
$r^3+4r^2+r-6=0\\$
$(r-1)(r^2+5r+6)=0\\$
$r=1, r=-2, r=-3\\$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}\\$
$W=\begin{vmatrix}
e^t & e^{-2t} & e^{-3t}\\
e^t & -2e^{-2t} & -3e^{-3t}\\
e^t & 4e^{-2t} & 9e^{-3t}\\
\end{vmatrix}=-12e^{-4t}\\$
$\text{$\therefore c=-12$ }\\$
$\text(c)\\$
$y_p(t)=Ate^t\\$
$y_p^{\prime}(t)=Ae^t+Ate^t\\$
$y_p^{\prime\prime}(t)=2Ae^t+Ate^t\\$
$y_p^{\prime\prime\prime}(t)=3Ae^t+Ate^t\\$
$3Ae^t+Ate^t+8Ae^t+4Ate^t+Ae^t+Ate^t-6Ate^t=12Ae^t\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad 12A=24\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad A=2$
$\text{We finally have the general solution,}\\ y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$
$\text{Using abels theorem,}
\\
W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}\\$
$\text(b)\\$
$\text{We have the following characteristic polynomial,}\\$
$r^3+4r^2+r-6=0\\$
$(r-1)(r^2+5r+6)=0\\$
$r=1, r=-2, r=-3\\$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}\\$
$W=\begin{vmatrix}
e^t & e^{-2t} & e^{-3t}\\
e^t & -2e^{-2t} & -3e^{-3t}\\
e^t & 4e^{-2t} & 9e^{-3t}\\
\end{vmatrix}=-12e^{-4t}\\$
$\text{$\therefore c=-12$ }\\$
$\text(c)\\$
$y_p(t)=Ate^t\\$
$y_p^{\prime}(t)=Ae^t+Ate^t\\$
$y_p^{\prime\prime}(t)=2Ae^t+Ate^t\\$
$y_p^{\prime\prime\prime}(t)=3Ae^t+Ate^t\\$
$3Ae^t+Ate^t+8Ae^t+4Ate^t+Ae^t+Ate^t-6Ate^t=12Ae^t\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad 12A=24\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad A=2$
$\text{We finally have the general solution,}\\ y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$