How to Guess the Particular Solution for y'' - 2y' + y = te^t?

[unseenoi]

hello can someone explain how to guess the yp for this Non-homogeneous differential equation

y'' - 2y' + y = te^t

characteristic polynomial: (y - 1)^2 so the characteristic roots are: y1=y2= 1

c1 and c2 are constant

for yh = (c1)e^t + (c2)te^t

please explained how to guess for te^t

 

[HallsofIvy]

I wouldn't call it "guessing"- there are rules you follow.

Any time you have a polynomial or power of t on the right side, try the polynomial up to the highest power. For t you would try At+ B. Here you have te^t so try (At+ B)e^t.

That's the basic rule. However, we also have the rule that when a possible particular solution is already a solution to the homogeneous solution, we must multiply by t to get something new.

Here, both e^t and te^t are solutions to the homogeneous solution. Multiplying (At+ B)e^t by t would give us (At^2+ Bt)e^t but since that still contains te^t which is a solution to homogeneous equation, we multiply by t again: try (At^3+ Bt^2)e^t.

 

[unseenoi]

My professor uses this as her particular solution: yp = At^3e^t but I don't understand why can someone help me?

 

[itr]

your professor is using the method of undetermined coefficients, you will use that, find your y' and y" and plug it into the origional equation, then solve for A, then that is your yp and you add it to your yh and you got your final answer