Diff EQ: Method of Undetermined Coefficients

In summary, the problem requires solving using the Method of Undetermined Coefficients. The homogeneous solution is obtained using the characteristic equation. For the particular solution, it is assumed that it takes the form of a linear combination of each forcing function in f(t). In this case, since the forcing function is polynomial, it does not matter how it is broken up. The solution can be obtained by testing different types of particular solutions if the forcing function has different classes of functions. A possible error in the solution may be due to an algebraic error rather than the assumed f's.
  • #1
Saladsamurai
3,020
7

Homework Statement



Solve using Method of Undetermined Coefficients:

y'' - 2y' + 3y = 15t - 8

The Attempt at a Solution



First I used the characteristic equation to solve for the homogeneous solution to get:

yh = C1e-t + C2e3t

Great. Now I need to find the particular. I am suppose to assume that each particular solution takes the form of a linear combination of each forcing function in f(t) on the right hand side of the original DE: 15t - 8.

Now here is my question (it's kind of silly):

Do I use:

yp1 = A*(15t - 8) +B*(d/dt)[15t - 8] and
yp2 = 0

or

yp1 = A*15t +B*(d/dt)[15t] and
yp2 = C(8)

or does it matter? I feel like it should not matter. My book says that the right hand side (the forcing function) of a second order DE takes the form f(t) = f1(t) + f2(t)

When the terms are distinct as in f(t) = t3 + sin(t) it is easy to distinguish the f's --> f1 = t3 and f2 = sin(t).

But when it is polynomial: f(t) = at3 + bt2 + ct +d
I feel like it should not matter how I break up f(t).

Any thoughts?
 
Physics news on Phys.org
  • #2
You're correct. You only need to test different types of particular solutions if the forcing function has different classes of functions.
 
  • #3
fzero said:
You're correct. You only need to test different types of particular solutions if the forcing function has different classes of functions.

Ok great! Thank you :smile: I was trying to figure out if a solution I was getting was because of my assumed f's (e.g. the ones I made above) or because of an algebraic error. I am pretty sure I can chalk it up to the latter.

Thanks again.
 

1. What is the Method of Undetermined Coefficients?

The Method of Undetermined Coefficients is a technique used to solve non-homogeneous linear differential equations. It involves finding a particular solution for the equation by assuming a form for the solution and then solving for the coefficients.

2. When is the Method of Undetermined Coefficients used?

The Method of Undetermined Coefficients is typically used when the non-homogeneous term in a differential equation is a polynomial, exponential, or trigonometric function.

3. How does the Method of Undetermined Coefficients work?

The method involves two main steps: first, assuming a particular form for the solution, and second, substituting this solution into the original differential equation and solving for the coefficients. These coefficients are then used to construct the final particular solution.

4. What is the difference between the Method of Undetermined Coefficients and the Method of Variation of Parameters?

The Method of Undetermined Coefficients assumes a specific form for the solution and solves for the coefficients, while the Method of Variation of Parameters involves finding a general solution and using a variation of it to find a particular solution.

5. Are there any limitations to using the Method of Undetermined Coefficients?

Yes, the method is limited to non-homogeneous linear differential equations with specific types of non-homogeneous terms. It also may not work for all cases, in which case the Method of Variation of Parameters or other techniques may need to be used.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
501
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
992
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Replies
4
Views
2K
  • Differential Equations
Replies
7
Views
1K
Back
Top