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MAT2442018F => MAT244Tests => Term Test 2 => Topic started by: Victor Ivrii on November 20, 2018, 05:51:00 AM

Consider equation
\begin{equation}
y'''2y'' y'+2y= 8e^{t}.
\label{21}
\end{equation}
(a)
Write a differential equation for the 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 solution of (\ref{21}).

This is my solution, thanks!

Here we can use Abel's Theorem to determine the value of the Wronskian
$$ \mbox{Therefore, } W = c e^{2t}, since p(t) = 2 $$
for part b, we must find the characteristic equation
$$ r^3  2r^2  r + 2 = 0 $$
$$ (r1)(r+1)(r2) = 0 $$
$$ \mbox{Therefore, } r = 1, 1, 2 $$
$$ \mbox{Therefore, } y_1(t) = e^{t} y_2(t) = e^{t} y_3(t) = e^{2t} $$
$$ \mbox{Therefore, the Wronskian} W(y_1, y_2, y_3)(t) = \begin{bmatrix} e^{t}&e^{t}&e^{2t}\\ e^{t}&e^{t}&2e^{2t}\\e^{t}&e^{t}&4e^{2t}\\ \end{bmatrix} $$
$$ det(\begin{bmatrix} e^{t}&e^{t}&e^{2t}\\ e^{t}&e^{t}&2e^{2t}\\e^{t}&e^{t}&4e^{2t}\\ \end{bmatrix}) = 6 e^{2t} $$
This is similiar to the solution found in part a, except that c = 6
for part c,
$$ \mbox{Assume} y(t) = Ate^{t} $$
$$ \mbox{Therefore, } y'(t) = Ae^{t}(1+t) $$
$$ \mbox{Therefore, } y''(t) = Ae^{t}(2+t) $$
$$ \mbox{Therefore, } y'''(t) = Ae^{t}(3+t) $$
Plugging this in the original equation
$$ Ae^{t}(t + 3  4 2t  1  t  2t) = 8Ae^{t} $$
$$ \mbox{Therefore, A = 4} $$
$$ \mbox{Therefore, the general equation is } y(t) = c_1e^{t} + c_2e^{t} + c_3e^{2t} 4te^{t} $$