## Derivitives - Superposition (attempted solution not correct)

1. The problem statement, all variables and given/known data

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 im going wrong and how I can correct it.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
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 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus 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.

 Quote by vela 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

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