Derivitives - Superposition (attempted solution not correct)

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Homework Help Overview

The problem involves solving a linear differential equation using the superposition method, specifically addressing the equation y'' + 6y' + 8y = 6sin(3t). The original poster attempts to find both the homogeneous and particular solutions but expresses uncertainty about the correctness of their approach, particularly regarding the sine term in the particular solution.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of including both sine and cosine terms in the particular solution, questioning the original poster's omission of the cosine term. There is an exploration of the correct form for the particular solution and the reasoning behind it.

Discussion Status

Some participants have provided guidance on the structure of the solution, emphasizing the need for both components in the particular solution. The original poster acknowledges the need for correction but continues to struggle with determining the proper sine term.

Contextual Notes

The original poster indicates a misunderstanding of the procedure involved in solving the differential equation, particularly in relation to the superposition method and the roles of the homogeneous and particular solutions.

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Homework Statement



Use the super position method to find the solution of:

y"+6y'+8y=6sin3t


2. The attempt at a solution

x^2+6x+8=6sin3t

found the x values x= -2,-4

yc=Asin3t+Bcos3t
y'=3Acos3t-3Bsin3t
y"=-9Asin3t-9Bcos3t

sin3t values (8A,-18B,-9A) A=18B-6
cos3t values(8B,18A,-9B) B=-0.018 ---------- A=-6.33

c1e^-2t+c2e^-4t-6.33sin3t-.018cos3t


I know that the sin term is not correct can someone explain where I am going wrong and how I can correct it.
 
Last edited:
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It seems like you're just following a procedure without understanding why you're doing what you're doing.

When you have a linear differential equation, the complete solution y(t) consists of a homogeneous part yh(t) and a particular part yp(t). The homogenous part satisfies the differential equation with the RHS set to zero:

y''h + 6y'h + 8yh = 0

It is this equation which you solve using the associated polynomial equation

x2 + 6x + 8 = 0

which you can solve to find the roots x=-2 and -4, which yields the homogeneous solution yh(t)=c1e-2t+c2e-4t. Those values aren't solutions to

x2 + 6x + 8 = 6 sin 3t

as you wrote.

To find the particular solution yp(t), you look at the forcing function. Here, you have a sine term which doesn't appear as part of the homogeneous solution, so yp(t) will have the form

yp(t)=A sin 3t + B cos 3t

You need both the sine and cosine terms to find the correct solution. Your mistake was leaving out the cosine term. Try again with the new trial solution.
 
vela said:
You need both the sine and cosine terms to find the correct solution. Your mistake was leaving out the cosine term. Try again with the new trial solution.

I redid it with the cosine function but still can determine the proper sine term at all i understand that it is the two parts
 
What?