Second Order Non-homogeneous Constant Coefficient Differential Equation

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Homework Help Overview

The discussion revolves around finding a general solution to a second-order non-homogeneous constant coefficient differential equation. The equation includes a polynomial and an exponential term on the right-hand side.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the form of the particular solution, with one suggesting a polynomial guess of degree higher than the inhomogeneous part. Questions arise regarding the inclusion of coefficients and the impact of existing roots from the complementary solution.

Discussion Status

There is an ongoing exploration of the appropriate form for the particular solution, with some guidance provided on adjusting the degree of the polynomial guess. Participants are actively engaging with the problem, and there is a recognition of the need to modify their approaches based on the characteristics of the differential equation.

Contextual Notes

Participants note the presence of characteristic roots in the homogeneous solution, which influences their choices for the particular solution. There is also mention of constraints related to the coefficients in the proposed solution forms.

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Homework Statement


Find a general solution to \frac{d^2x}{dt^2}-2\frac{dx}{dt}=1-4t+e^t


Homework Equations


None really.


The Attempt at a Solution


I know that a complimentary solution is x=c_1+c_2e^{2t}
But when I try to guess say: x_p=At+B+Ce^t and plug into the DE, I do not get anything to equate to 4t. Do I have to guess a degree higher? And if so, do I include all coefficients such that my guess becomes x_p=At^2+Bt+C+De^t?
 
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You're right, with a polynomial in the inhomogeneous part of your equation, you should check a degree higher than it. You should include all the coefficients as well, but it turns out C has no constraints (no x(t) part of your differential equation on the L.S.) so you can set it to 0. The rest comes out easily after substituting back into the DE.
 
"At+ B" corresponds to characteristic root 0 which is already a characteristic root of your homogeneous equation. Try y= At^2+ Bt+ Ce^t instead.
 
Great thanks! Worked it out. So I guess I'm going by the fact that because my complementary solution has a constant term, I just multiply (At+B) by t.
 

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