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anonymous
 5 years ago
I need to find the particular solution to u"5u'+4u=2e^t but when I plug in e^t it comes out 0=2 I don't know what went wrong.
anonymous
 5 years ago
I need to find the particular solution to u"5u'+4u=2e^t but when I plug in e^t it comes out 0=2 I don't know what went wrong.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since your characteristic equation is: \[Yc=C_{1}e^{4t}+C_{2}e^{t}\] and contains the form \[Ae^t \rightarrow C_{2}e^{t}\] you cant use \[Ae^t\] you have to multiply \[Ae^t\] by t to get \[Ate^t\] and then solve using the method of undetermined coefficients \[Y=Ate^t,Y^{'}=Ae^t+Ate^t, Y^{''}=2Ae^t+Ate^t\] Plug these guys in to the differential equation \[2Ae^t+Ate^t5Ae^t5Ate^t+4Ate^t=2e^t\] Resulting in: \[3Ae^t=2e^t \rightarrow A=2/3\] \[Y(t)=C_{1}e^{4t}+C_{2}e^t2/3te^t\] hope this helps
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