### Author Topic: TUT 0602 Quiz 4  (Read 1743 times)

#### Yichen Ji

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##### TUT 0602 Quiz 4
« on: November 04, 2019, 05:01:07 PM »
Question:find the general solution of the equation:

y''+y'-6y=12e^{3t}+12e^{-2t}

Solution:
First consider homogeneous equation $y''+y'-6y=0$
the characteristic polynomial equation is
\begin{align*}
r^2+r-6 &=0\\
(r+3)(r-2) &=0\\
\end{align*}
We get $r_1=-3$ and $r_2=2$
Therefore, the homogeneous solution is $y_h(t)=c_1e^{-3t}+c_2e^{2t}$
Next,we look for $y_p(t)$
Consider
\begin{align*}
y_1(t) &=Ae^{3t}\\
y_1'(t) &=3Ae^{3t}\\
y_1''(t) &=9Ae^{3t}\\
6Ae^{3t} &=12e^{3t}\\
A &= 2
\end{align*}
So $y_1(t)=2e^{3t}$
Then consider
\begin{align*}
y_2(t) &=Be^{-2t}\\
y_2'(t) &=-2Be^{-2t}\\
y_2''(t) &=4Be^{-2t}\\
-25Be^{-2t} &=12e^{-2t}\\
B &=-\frac{12}{25}\\
\end{align*}
So $y_2(t)=-\frac{12}{25}e^{-2t}$
And the particular solution is
\begin{align*}
y_p(t) &=y_1(t)+y_2(t)\\
y_p(t) &=2e^{3t}-\frac{12}{25}e^{-2t}\\
\end{align*}
To conclude, the general solution is
\begin{align*}
y_g(t) &=y_h(t)+y_p(t)\\
y_g(t) &=c_1e^{-3t}+c_2e^{2t}+2e^{3t}-\frac{12}{25}e^{-2t}\\
\end{align*}
« Last Edit: November 04, 2019, 06:05:17 PM by Yichen Ji »