Derivitives - Superposition (attempted solution not correct)

In summary: You're still trying to find the sine term even though you already found it? What is the point of this exercise?
  • #1
kylerawn
8
0

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.
 
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  • #2
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.
 
  • #3
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
 
  • #4
What?
 

FAQ: Derivitives - Superposition (attempted solution not correct)

What is the concept of superposition in derivatives?

The concept of superposition in derivatives refers to the principle that allows us to combine multiple derivatives to find the derivative of a more complex function. This is possible because derivatives are linear operators, meaning they can be added and subtracted from each other.

How do you solve for derivatives using superposition?

To solve for derivatives using superposition, you first need to break down the function into simpler parts. Then, you can use the basic derivative rules to find the derivative of each part. Finally, you can combine the individual derivatives using addition or subtraction to find the overall derivative of the function.

Can superposition be used to find higher-order derivatives?

Yes, superposition can be used to find higher-order derivatives. The process is similar to finding first-order derivatives, but you may need to apply the derivative rules multiple times depending on the complexity of the function.

What are some common mistakes when using superposition to find derivatives?

One common mistake is not properly breaking down the function into simpler parts, making it difficult to find the individual derivatives. Another mistake is not applying the derivative rules correctly, leading to incorrect results. It is also important to keep track of negative signs and constant terms when combining the individual derivatives.

Can superposition be used to find derivatives of both continuous and discontinuous functions?

Yes, superposition can be used to find derivatives of both continuous and discontinuous functions. However, for discontinuous functions, you may need to use the piecewise function approach and apply the derivative rules to each piece separately before combining them using superposition.

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