How Do You Solve a Complex IVP with Sinusoidal and Polynomial Terms?

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Discussion Overview

The discussion revolves around solving a complex initial value problem (IVP) involving a second-order linear differential equation with both sinusoidal and polynomial terms. Participants explore various methods for finding particular solutions, including undetermined coefficients and variation of parameters, while also addressing challenges encountered in their calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents their approach using undetermined coefficients, proposing a particular solution of the form At^2 + Bt + C + Dcos(2t) + Esin(2t) but expresses uncertainty about the correctness of their coefficients.
  • Another participant questions the dependency of A and B on t, suggesting they should be constants, and provides their own particular solution as y = t^2 - 6t + 14 - 6cos(2t) - 7sin(2t.
  • A later reply acknowledges the error in assuming A and B depend on t and corrects their approach, confirming A=1 based on the comparison with the polynomial term.
  • One participant suggests that variation of parameters should be worked out in full, especially for those new to the method, and mentions the potential use of Laplace transforms.
  • Another participant discusses the use of reduction of order and the choice of functions for integration, indicating that using y1 may lead to more cancellations.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to finding the particular solution, with some advocating for undetermined coefficients while others suggest variation of parameters. There is no consensus on the best method or the correctness of the solutions presented.

Contextual Notes

Some participants note the complexity of the calculations involved, particularly with the variation of parameters method, and the potential for errors in deriving coefficients. There are also mentions of specific assumptions and dependencies that may affect the outcomes.

Who May Find This Useful

This discussion may be useful for students and practitioners dealing with second-order linear differential equations, particularly those interested in methods for solving initial value problems involving sinusoidal and polynomial terms.

cronxeh
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Ok I'm stuck on this problem - and it didnt particularly look hard when I started doing it, oh say about a good 2 hours ago.. hmm

Solve the IVP
[tex]2y'' + 3y' + y = t^2 + 85 sin(2t), y(0) = -2, y'(0) = 0.[/tex]

My homogeneous solution is [tex]c1y1 + c2y2[/tex]

[tex]y1 = e^\frac{-t}{2}, y2 = e^-^t, Wronskian = e^\frac{-3t}{2}+e^\frac{-t}{2}[/tex]

So for a particular solution I've tried 2 different methods

1. Undetermined Coefficients
Assumed solution was [tex]At^2 + Bt + C + Dcos(2t) + Esin(2t)[/tex]

[tex]yp' = 2At + B - 2Dsin(2t) + 2Ecos(2t)[/tex]
[tex]yp'' = 2A - 4Dcos(2t) -4Esin(2t)[/tex]

After pluggin this back into original equation [tex]2y'' + 3y' + y[/tex] I got the following equalities:
[tex]At^2 + Bt + C + 6At + 3B + 4A = t^2[/tex]
[tex]cos(2t)(-7D + 6F) = 0[/tex]
[tex]sin(2t)(-7F - 6D) = 85sin(2t)[/tex]

Fiddling around with the equalities I got C=0, D=-6, F=-7

[tex]A=\frac{3t}{3t+14}, B=\frac{-4t}{3t+14}[/tex]

After assembling the whole thing together I don't get the right hand side after integrating to check the solutions, and so even when I derive the y=yp+yh solution to get my constants for homogeneous equation I get wrong coefficients c1 and c2 for yh - an infinite number of them, so I'm pretty sure my solution is wrong

Should I have multiplied the assumed solution by t anywhere?

2. Variation of Parameters

[tex]Yp = -y1 \int \frac{y2 g(t) dt}{Wronskian} + y2 \int \frac{y1 g(t) dt}{Wronskian}[/tex]
where [tex]g(t) =\frac{ t^2 + 85sin(2t) } {2}[/tex]
The result was extremely long and contained a lot of exponentials, and I assumed it to be incorrect, although any comment on the correct answer you got would be appreciated
 
Last edited:
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In your undetermined coefficients solution, why do A and B depend on t? A and B are supposed to be constants, right?

The particular solution I get is:

[tex]y = t^2 - 6t + 14 -6\cos 2t -7 \sin 2t[/tex]
 
cronxeh said:
Ok I'm stuck on this problem - and it didnt particularly look hard when I started doing it, oh say about a good 2 hours ago.. hmm

Solve the IVP
[tex]2y'' + 3y' + y = t^2 + 85 sin(2t), y(0) = -2, y'(0) = 0.[/tex]

My homogeneous solution is [tex]c1y1 + c2y2[/tex]

[tex]y1 = e^\frac{-t}{2}, y2 = e^-^t, Wronskian = e^\frac{-3t}{2}+e^\frac{-t}{2}[/tex]

So for a particular solution I've tried 2 different methods

1. Undetermined Coefficients
Assumed solution was [tex]At^2 + Bt + C + Dcos(2t) + Esin(2t)[/tex]

[tex]yp' = 2At + B - 2Dsin(2t) + 2Ecos(2t)[/tex]
[tex]yp'' = 2A - 4Dcos(2t) -4Esin(2t)[/tex]

I get the following equations:

4A+3B+C=0
6A+B=0
A=1
6E-7D=0
-7E-6D=85

Edit: Suppose I should compare this to the results reported by James . . .
Ok, I'm cool.
 
Last edited:
James R said:
In your undetermined coefficients solution, why do A and B depend on t? A and B are supposed to be constants, right?

The particular solution I get is:

[tex]y = t^2 - 6t + 14 -6\cos 2t -7 \sin 2t[/tex]


Doh.. of course! So since [tex]At^2 = t^2[/tex] then [tex]A=1[/tex] and I simply equate the other terms to 0. Thanks.
 
With variation of parameters you typically want to work out the entire step, especially if you're new to it, I prefer the practical variation of parameters approach v.s. plugging into the wronskian derivation. Reduction of order should work with this also, except try using your y1 intstead of the simpler y2. Have you gotten to the laplace transform yet?
 
I think that you will have more cancellations with the y1, however if you need to go with y2 (using reduction of order), you'll probably need to use euler's form of sin(2t) to solve for your first integral
 

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