Author Topic: TUT 0602 Quiz 3  (Read 500 times)

Yichen Ji

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TUT 0602 Quiz 3
« on: November 04, 2019, 05:36:53 PM »
Question:find a differential equation whose solution is $y=c_1e^{2t}+c_2e^{-3t}$
Solution:
Set
\begin{align*}
    y_1 &=e^{2t} & y_2 &=e^{-3t}\\
    y_1' &=2e^{2t} & y_2' &=-3e^{-3t}\\
    r_1 &=2 & r_2 &=-3\\
\end{align*}
So the Wronskian
\begin{equation*}
    \begin{split}
   W[y_1,y_2](t) &= \begin{vmatrix}
y_1(t)&y_2(t)\\
y_1'(t)&y_2'(t)
\end{vmatrix} \\
&=y_1y_2'-y_2y_1' \\
&=-5e^{-t}\\
&\neq0
\end{split}
\end{equation*}
Then ${y_1,y_2}$forms a fundamental set of solution.We can write one characteristic equation as
\begin{align*}
    (r-2)(r+3)=0\\
    (r-2)(r+3)e^{rt}=0
\end{align*}
by multiplying $e^{rt}$on both sides,we get one equation for this general solution:
\begin{equation*}
    y''+y'-6y=0
\end{equation*}
« Last Edit: November 04, 2019, 06:04:38 PM by Yichen Ji »