Hello, I've got an undetermined coefficients question. The problem is y''-3y'-4y=e^4t. If I use Ae^4t as my likely sol, A just dissapears

Question

badhi · Answer

cant you say that for all non zero values of A it satisfies the equation?

dumbcow · Answer

that means your guess is wrong ... try Ate^4t

anonymous · Answer

You need to consider higher powers in t, let : $$y = A t e^{4t}+Be^{4t}$$so
$$y^{'}=A(e^{4t}+4t e^{4t})+4Be^{4t}$$$$=(4A) te^{4t}+(A+4B)e^{4t}$$and$$y^{''}=4A(e^{4t} + 4te^{4t}) + 4(A+4B)e^{4t}$$$$=(16A)te^{4t}+(8A+16B)e^{4t}$$so$$y^{''}-3y^{'}-4y=(16A - 12A -4A)te^{4t}+(8A+16B-3A-12B-4B )e^{4t}$$$$=(0A)te^{4t}+(5A)e^{4t}=(1)e^{4t}$$this tells you that B wasn't actually needed, but$$5A = 1$$or$$A= \frac{1}{5}$$which you can check by substitution ie. $$y=(1/5)te^{4t}$$