Author Topic: TUT 0202 QUIZ3  (Read 264 times)

Yijin Qiang

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TUT 0202 QUIZ3
« on: October 11, 2019, 02:00:01 PM »
$we\,asume\,that\,y=e^{rt},and\,then\,it\,follows\,that\,r\,must\,be\,a\,root\,of\,characteristic\\
2r^{2}-3r+1=(2r-1)(r-1)=0\\
Hence\\
\begin{Bmatrix}
r_{1=\frac{1}{2}}\\r_{2}=1

\end{Bmatrix}\\
Since\,the\, general \,solution\, has \,the\,form\,of\\
y=c_{1}e^{r_{1}t}+c_{2}e^{r{2t}}\\
therefore,\,the\,general\,solution\,of\,the\,given\,differential\,eqution\,is\\
y=c_{1}e^{\frac{1}{2}t}+c_{2}e^{t}$
« Last Edit: October 11, 2019, 02:08:15 PM by Yijin Qiang »