Undetermined Coefficients to solve

In summary, Undetermined Coefficients is a method used to solve differential equations by setting up a system of equations and determining the coefficients for the particular solution. In this conversation, the speaker sets up the system and gets an answer of y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx, while the book's answer is y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx-1/4x^2. The difference lies in the inclusion of a term to account for the constant on the RHS of the differential equation. After correcting this, the speaker confirms that the book's answer is correct.
  • #1
kuahji
394
2
Undetermined Coefficients to solve.

y'''-6y''=3-cosx

So I set everything up, get my system & I get an answer of

y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx

the book has y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx-1/4x^2.

So I'm not sure where the book gets the last term, any ideas?
 
Physics news on Phys.org
  • #2
The book's answer is correct, but I have no idea where you are going wrong since you haven't shown your work...
 
  • #3
Ok, I have yp= A + B cosx + c sinx
y'p=-Bsinx+C cosx
y''p=-B cosx - C sinx
y'''P=Bsinx - C cosx

After I have plugged these back into the LHS, I get a system of
B+6C=0
6B-C=-1

These give me my coefficients -6/37cosx & 1/37sinx.
 
  • #4
Oh & yh=r^2(r-6), which is where I get c1+c2x+c3e^6c.
 
  • #5
kuahji said:
Ok, I have yp= A + B cosx + c sinx

Well that's your problem right there, the constant term 'A' is not linearly independent to your homogeneous solution, which also contains a constant term... to account for the constant term (3) on the RHS of your DE, you need to add a term of the form Ax^r where r is the smallest non-negative integer such that no terms of the same order are present in your homogeneous solution...

kuahji said:
where I get c1+c2x+c3e^6x
 
  • #6
Thanks. It works out nicely now.
 

1. What is the method of undetermined coefficients used for?

The method of undetermined coefficients is a technique used to find a particular solution to a non-homogeneous linear differential equation. It is particularly useful for solving equations with constant coefficients.

2. How does the method of undetermined coefficients work?

The method of undetermined coefficients works by assuming a particular form for the solution and plugging it into the differential equation. The coefficients of the assumed solution are then determined through substitution and comparison with the original equation.

3. What are the requirements for using the method of undetermined coefficients?

The method of undetermined coefficients can only be used for linear differential equations with constant coefficients and non-homogeneous terms. The non-homogeneous term must also be in the form of a polynomial, exponential, sine, or cosine function.

4. What is the difference between the method of undetermined coefficients and variation of parameters?

The method of undetermined coefficients is a specific technique used to solve a particular type of non-homogeneous linear differential equation. Variation of parameters, on the other hand, is a more general method that can be used to solve non-homogeneous equations with variable coefficients.

5. Are there any limitations or drawbacks to using the method of undetermined coefficients?

Yes, the method of undetermined coefficients can only be used for a limited set of non-homogeneous equations. It also does not work for equations with repeated roots, in which case another method such as variation of parameters must be used. Additionally, the method may not always yield a solution and may require additional techniques to find a complete solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
491
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
692
  • Calculus and Beyond Homework Help
Replies
3
Views
986
  • Calculus and Beyond Homework Help
Replies
7
Views
772
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Differential Equations
Replies
16
Views
3K
Replies
5
Views
2K
Back
Top