Question

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.

Answer 1

since 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
\[2Ae^t+Ate^t-5Ae^t-5Ate^t+4Ate^t=2e^t\]
Resulting in:
\[-3Ae^t=2e^t \rightarrow A=-2/3\]
\[Y(t)=C_{1}e^{4t}+C_{2}e^t-2/3te^t\]
hope this helps