1. ## undetermined coefficients

I'm supposed to find the solution this initial value problem using the method of undetermined coefficients
$
y'' - 2y' - 3y = 3te^{2t}, y(0)=1, y'(0)=0.$

My guess was to use $Y(t) = Ate^{2t} + Be^{2t}$ for the particular solution, but this led me to the wrong answer.
I have a hunch that I should use $Y(t) = t(Ate^{2t} + Be^{2t})$ but I have no idea why, since $e^{2t}$ is not part of the complimentary solution.

I appreciate any help.

2. the P.I. should be c1.e^-t+c2.e^3t
and the A.E.=3.e^2t(t-6)
are they correct???

we get the P.I. by solving the quadratic in y' ie. D^2-2D-3=0
which gives D=-1,3

and the A.E. is calculated by separating the exponential part then........(it should be given in any text book).hope this was helpful!

I'm supposed to find the solution this initial value problem using the method of undetermined coefficients
$
y'' - 2y' - 3y = 3te^{2t}, y(0)=1, y'(0)=0.$

My guess was to use $Y(t) = Ate^{2t} + Be^{2t}$ for the particular solution, but this led me to the wrong answer.
What did you do that gave you a wrong answer?

If $y= Ate^{2t}+ Be^{2t}$, then
$y"= Ae^{2t}+ 2Ate^{2t}+ 2Be^{2t}= (A+ 2B)e^{2t}+ 2Ate^{2t}$
$y"= 2(A+ 2B)e^{2t}+ 2Ae^{2t}+ 4Ate^{2t}= (6A+ 2B)e^{2t}+ 4Ate^{2t}$
Putting those into the equation gives
$[(6A+ 2B)-2(A+ 2B)- 3B]e^{2t}+ [4A- 4A- 3A]te^{2t}$ $= (4A- 5B)e^{2t}+ (-3A)te^{2t}= 3te^{2t}$

We must have 4A- 5B= 0 and -3A= 3. A= 1 so 4- 5B= 0 and B= 4/5.
That is a perfectly good solution.

I have a hunch that I should use $Y(t) = t(Ate^{2t} + Be^{2t})
$
but I have no idea why, since $e^{2t}$ is not part of the complimentary solution.

I appreciate any help.