2nd order differential equation: undetermined coefficients

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Homework Help Overview

The problem involves solving a second-order differential equation using the method of undetermined coefficients, specifically addressing the equation y'' - y = t - 4e^(-t).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the characteristic equation and the form of the particular solution. There is a focus on the implications of having e^(-t) as a solution in the complementary equation, leading to questions about modifying the guess for the particular solution.

Discussion Status

Some participants have offered guidance on adjusting the form of the particular solution to account for the existing solution in the complementary equation. There is an ongoing exploration of how to correctly substitute and derive coefficients.

Contextual Notes

Participants express uncertainty regarding the correct form of the particular solution and the implications of their previous attempts. There is acknowledgment of mistakes in derivative calculations, which may affect the progress of the discussion.

dmoney123
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Homework Statement



y''-y=t-4e^(-t)

Homework Equations



method of undetermined coefficients

The Attempt at a Solution



solving for characteristic equation first

y''-y=0

r^2-1=0

c_1e^(-t)+c_2e^(t)

RHS

particular solution

t-4e^(-t)

y_p(t)= At+B+Ce^(-t)

y_pt'(t)=A-Ce^(-t)

y_p''(t)=Ce^(-t)

plug into LHS

Ce^(-t)-At-B-Ce^(-t)=t-4e^(-t)

-A=t
A=-1

B=0

C cancels on the left... I am not sure how to any further?

Thanks
 
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dmoney123 said:

Homework Statement



y''-y=t-4e^(-t)

Homework Equations



method of undetermined coefficients

The Attempt at a Solution



solving for characteristic equation first

y''-y=0

r^2-1=0

c_1e^(-t)+c_2e^(t)

RHS

particular solution

t-4e^(-t)

y_p(t)= At+B+Ce^(-t)
No, this won't work, since e-t is already a solution in the complementary equation. Do you know what to do when you run into this?
dmoney123 said:
y_pt'(t)=A-Ce^(-t)

y_p''(t)=Ce^(-t)

plug into LHS

Ce^(-t)-At-B-Ce^(-t)=t-4e^(-t)

-A=t
A=-1

B=0

C cancels on the left... I am not sure how to any further?

Thanks
 
Mark44 said:
No, this won't work, since e-t is already a solution in the complementary equation. Do you know what to do when you run into this?

You add a coefficient t

so the guess becomes At+B+Cte^(-t) right? When I plug that back in, it still cancels out
 
dmoney123 said:
You add a coefficient t

so the guess becomes At+B+Cte^(-t) right? When I plug that back in, it still cancels out
I'm not sure what you mean.
If yp = At + B + Cte-t, then yp'' - yp has to equal t - 4e-t. Take yp and its second derivative, and substitute them into your nonhomogeneous equation to get A, B, and C.
 
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Got it... I was making a mistake with my derivative... I feel stupid.

Thanks Mark44!
 

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