Author Topic: MAT244 Higher Order Homogeneous Equation  (Read 277 times)

shaohan shi

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MAT244 Higher Order Homogeneous Equation
« on: November 27, 2018, 06:00:54 PM »
Consider equation $y^{'''} - 7y' + 6y = 100e^{-3t}$.
a. Write equation for Wronskian of $\{y_1, y_2, y_3\}$, which are solutions for homogeneous equation and solve it.
b. Find fundamental system $\{y_1, y_2, y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with(a).
c.Find the general soltuion of (1).

wenlinwang

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Re: MAT244 Higher Order Homogeneous Equation
« Reply #1 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 $y(t) = y_c(t) + y_p(t)$
$y(t) = c_1e^t +c_2e^{2t} + c_3e^{-3t} + 5te^{-3t}$