Is My Approach to Solving the 3rd Order IVP Correct?

Click For Summary

Discussion Overview

The discussion revolves around solving a third-order initial value problem (IVP) involving a differential equation. Participants explore various approaches to finding the complementary and particular solutions, addressing the correct forms to use based on the equation's structure.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes using At^2 and B*e^t*t as particular solutions for the IVP.
  • Another participant suggests a different form for the particular integral, recommending f(t) = At^4 + Bt^3 + Ct^2 + Dt + E + Fe^t.
  • A third participant clarifies that three independent solutions to the associated homogeneous equation are y1(t) = 1, y2(t) = e^t, and y3(t) = e^{2t}, and questions the initial participant's notation of "Y1" and "Y2."
  • This participant also notes that since t is already a solution, the proposed form for the particular solution should be At^2 + Bt, and for e^t, it should be Cte^t.
  • One participant reflects on their earlier confusion regarding variable notation and acknowledges the potential contribution of the third derivative of t^4 to the right-hand side of the equation.
  • The initial poster clarifies their notation, stating that y1 and y2 refer to the "right side" of the equation.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate forms for the particular solution, with no consensus reached on the correct approach. The discussion remains unresolved regarding the best method to solve the IVP.

Contextual Notes

Participants highlight the dependence on the specific forms of solutions based on the characteristics of the differential equation, indicating that assumptions about the solutions may vary.

newtomath
Messages
37
Reaction score
0
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?
 
Physics news on Phys.org
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:
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"!
 
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)
 
Thanks. sorry for the confusion, by y1 and y 2 I meant the "right side" of t( at^2 +bt + Cte^t)
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
3
Views
1K