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MAT244--Lectures & Home Assignments / Re: MAT244 Higher Order Homogeneous Equation
« on: November 27, 2018, 06:07:36 PM »
a) $y^{'''} + 0y^{''} - 7y' + 6y =100e^{-3t}$
p(t) = 0
$$W = ce^{-\int p(t)dt} = ce^{c_1} = C$$
b) $r^3 -7r + 6 = 0$
$(r-1)(r-2)(r-3) = 0$
so r =1 r=2 r=-3
so $y(t) =c_1e^t + c_2e^{2t} + c_3e^{-3t}$
\begin{equation*}
W = \begin{bmatrix}
e^t & e^{2t} & e^{-3t}\\
e^t & 2e^{2t} & -3e^{-3t}\\
e^t & 4e^{2t} & 9e^{-3t}
\end{bmatrix} = 20
\end{equation*}
so W = C = 20
c) $y_p(t) = Ae^{-3t}$ same with home part
so $y_p(t) = Ate^{-3t}$
$y_p'(t) = Ae^{-3t} - 3Ate^{-3t}$
$y_p''(t) = 9Ate^{-3t} - 6Ae^{-3t}$
$y_p'''(t) = 27Ae^{-3t} - 27Ate^{-3t}$
$y^{'''} + 0y^{''} - 7y' + 6y =100e^{-3t}$
$27Ae^{-3t} - 27Ate^{-3t} -7Ae^{-3t} + 21Ae^{-3t} + 6Ate^{-3t} = 100e^{-3t}$
so A = 5
so $y_p(t) = 5te^{-3t}$
so $y(t) = y_c(t) + y_p(t)$
$y(t) = c_1e^t +c_2e^{2t} + c_3e^{-3t} + 5te^{-3t}$
p(t) = 0
$$W = ce^{-\int p(t)dt} = ce^{c_1} = C$$
b) $r^3 -7r + 6 = 0$
$(r-1)(r-2)(r-3) = 0$
so r =1 r=2 r=-3
so $y(t) =c_1e^t + c_2e^{2t} + c_3e^{-3t}$
\begin{equation*}
W = \begin{bmatrix}
e^t & e^{2t} & e^{-3t}\\
e^t & 2e^{2t} & -3e^{-3t}\\
e^t & 4e^{2t} & 9e^{-3t}
\end{bmatrix} = 20
\end{equation*}
so W = C = 20
c) $y_p(t) = Ae^{-3t}$ same with home part
so $y_p(t) = Ate^{-3t}$
$y_p'(t) = Ae^{-3t} - 3Ate^{-3t}$
$y_p''(t) = 9Ate^{-3t} - 6Ae^{-3t}$
$y_p'''(t) = 27Ae^{-3t} - 27Ate^{-3t}$
$y^{'''} + 0y^{''} - 7y' + 6y =100e^{-3t}$
$27Ae^{-3t} - 27Ate^{-3t} -7Ae^{-3t} + 21Ae^{-3t} + 6Ate^{-3t} = 100e^{-3t}$
so A = 5
so $y_p(t) = 5te^{-3t}$
so $y(t) = y_c(t) + y_p(t)$
$y(t) = c_1e^t +c_2e^{2t} + c_3e^{-3t} + 5te^{-3t}$