Derivitives - Superposition (attempted solution not correct)

kylerawn

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

vela

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 y_h(t) and a particular part y_p(t). The homogenous part satisfies the differential equation with the RHS set to zero:

y''_h + 6y'_h + 8y_h = 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 y_h(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 y_p(t), you look at the forcing function. Here, you have a sine term which doesn't appear as part of the homogeneous solution, so y_p(t) will have the form

y_p(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.

kylerawn

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

vela

What?