# Initial Value Problem

1. Jul 18, 2011

### newtomath

a 3rd order IVP I am havin trouble with:

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. Jul 19, 2011

### hunt_mat

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

Last edited: Jul 19, 2011
3. Jul 19, 2011

### HallsofIvy

I have no clue what you mean by "Y1" and "Y2". Three independent solutions to the associated homogeneous equation are y1(t)=1, $y2(t)= e^t$, and $y3(t)= e^{2t}$.

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

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. Jul 19, 2011

### hunt_mat

I was thinking that the third derivative of t^4 would contribute to the t term on the RHS. I got my x's and t's mixed up and i have now corrected it. It should all come out in the wash anyway (I think, it's been some years since I looked at equations such as these)

5. Jul 20, 2011

### newtomath

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