Derivitives - Superposition (attempted solution not correct)

[kylerawn]

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.

 

[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]

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

 

[vela]

What?