## 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.

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