- #1

- 8

- 0

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

- Thread starter unseenoi
- Start date

- #1

- 8

- 0

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

- #2

HallsofIvy

Science Advisor

Homework Helper

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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 [itex]te^t[/itex] so try [itex](At+ B)e^t[/itex].

That's the

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

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