- #1

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y''' -3y'' +2y' = t + e^t y(0)=1, y'(0)= -.25 y''(0)= -1.5

I am using At^2 and B*e^t *t as my Y1 and Y2. Is this correct?

- Thread starter newtomath
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- #1

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y''' -3y'' +2y' = t + e^t y(0)=1, y'(0)= -.25 y''(0)= -1.5

I am using At^2 and B*e^t *t as my Y1 and Y2. Is this correct?

- #2

hunt_mat

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I think that you were helped with the complimentary solution, for the particular integral I would try the function [itex]f(t)=At^{4}+Bt^{3}+Ct^{2}+Dt +E +Fe^{t}[/itex]

Last edited:

- #3

HallsofIvy

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Normally, with a "right side" of t, you would try [itex]At+ B[/itex] but since t is already a solution, you should try [itex]At^2+ Bt[/itex]. Normally with [itex]e^t[/itex] on the right side, you sould try [itex]Ce^t[/itex] but since [itex]e^t[/itex] is already a solution, you should try [itex]Cte^t[/itex].

hunt_mat usually gives very good responses but he may have been overly sleepy here. I can see no reason to include third or fourth power and certainly no reason to combine "x" and "t"!

- #4

hunt_mat

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- #5

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Thanks. sorry for the confusion, by y1 and y 2 I meant the "right side" of t( at^2 +bt + Cte^t)

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