Author Topic: TUT0401 Question  (Read 488 times)

Kexin Li

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TUT0401 Question
« on: October 18, 2019, 12:37:27 AM »
$$
Find\;the\;solution\;of\;the\;problem\\
2y'' - 3y' +y =0,y(0)=2,y'(0)=\frac{1}{2}\\
Assume \;that\; y=e^{rt}\\
Now\;y=e^{rt}\\
then\;y'=re^{rt}\\
and\;y=r^{2}e^{rt}\\
2r^{2}e^{rt}-3re^{rt}+e^{rt}=0\\
e^{rt}(2r^{2}-3r+1)\\
since\;e^{rt}\neq 0\\
then\;2r^{2}-3r+1=0\\
(r-1)(2r-1)=0\\
r=1,\frac{1}{2}\\
Then\;e^{t}and\;e^{\frac{1}{2}t}are\;two\;solutions\;of\;the\;equation\;and\;hence\;the\;general\;solution\;is\;given\;by\\
y=c_{1}e^{t}+c_{2}e^{\frac{t}{2}}\\
since\;y(0)=2\\
c_{1}e^{0}+c_{2}e^{0}=2\\
c_{1}+c_{2}=2\\
y'=c_{1}e^{t}+\frac{1}{2}c_{2}e^{\frac{t}{2}}\\
since\;y'(0)=\frac{1}{2}\\
c_{1}e^{0}+\frac{1}{2}c_{2}e^{0}=\frac{1}{2}\\
c_{1}+\frac{1}{2}c_{2}=\frac{1}{2}\\
c_{1}=-1,\;c_{2}=3\\
so\;the\;solution\;is\;y=3e^\frac{t}{2}-e^{t}
$$