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### Topics - Kexin Li

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

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##### Quiz-1 / TUT0401 Question
« on: September 27, 2019, 02:59:23 PM »
Find the general solution of the given differential equation, and use it to determine how solutions as t approaches infinity.
ty' +2y=sin(t), t>0

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