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Initial Value Problemby newtomath
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#1
Jul1811, 11:06 PM

P: 37

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
Jul1911, 06:57 AM

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P: 1,583

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]



#3
Jul1911, 07:29 AM

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P: 39,691

I have no clue what you mean by "Y1" and "Y2". Three independent solutions to the associated homogeneous equation are y1(t)=1, [itex]y2(t)= e^t[/itex], and [itex]y3(t)= e^{2t}[/itex].
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
Jul1911, 07:33 AM

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P: 1,583

Initial Value Problem
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
Jul2011, 02:11 PM

P: 37

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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