Author Topic: TUT5103 Quiz3  (Read 295 times)

Shang Wu

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TUT5103 Quiz3
« on: October 11, 2019, 01:56:28 PM »
Find the general solution of the given differential equation.
\begin{align*}
2y''-3y'+y=0
\end{align*}
Solution:
First we find the the characteristic equation, that is
\begin{align*}
2r^2-3r+1=0
\end{align*}
then we solve for the roots $r_1, r_2$
\begin{align*}
    r&=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\
    &=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(1)}}{2(2)} \\
    &= \frac{3 \pm 1}{4}
\end{align*}
So
\begin{align*}
r_1 &=\frac{3 + 1}{4}=1,\\
 r_2 &=\frac{3 - 1}{4}=\frac{1}{2}
\end{align*}
and
\begin{align*}
y_1 &= e^t, \\
y_2 &=e^{\frac{1}{2}t}
\end{align*}
Therefore we get the general solution
\begin{align*}
    y&=c_1y_1+c_2y_2\\
    &=c_1e^t + c_2e^{\frac{1}{2}t}
\end{align*}