Method of undetermined coefficients -- Help please

Click For Summary

Homework Help Overview

The problem involves using the method of undetermined coefficients to find the general solution of a second-order, linear, non-homogeneous differential equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the method of undetermined coefficients but expresses uncertainty about the next steps after deriving the characteristic equation and a particular solution form.

Discussion Status

Participants are providing guidance on notation and differentiation rules, with some pointing out potential oversights in the original poster's differentiation of the particular solution. Multiple aspects of the problem are being explored, including notation and the application of the product rule.

Contextual Notes

The original poster is new to the forum and seeks clarification on using superscripts for exponents, indicating a need for foundational understanding alongside the specific problem-solving process.

phantom lancer
Messages
2
Reaction score
0

Homework Statement


Using the method of undetermined coefficients, determine the general solution of the following second-order, linear, non-homogenous equations.

y'' - 4y' + 4y = 2e^(2x+3)

Homework Equations


I'm not sure what to do from here...
Also, I'm new here. How do I use the superscript for exponents?

The Attempt at a Solution


r^2 - 4r + 4 = 0
r = 2, so y = C1e^(2x) + C2xe^(2x)

I assume Yp = Ax^(2)e^(2x+3)
so, Yp' = 2Ax^(2)e^(2x+3)
Yp'' = 4Ax^(2)e^(2x+3)

plugging them in the equation: 4Ax^(2)e^(2x+3) - 4(2Ax^(2)e^(2x+3) + 4(Ax^(2)e^(2x+3) = Ax^(2)e^(2x+3)

I get 0 = Ax^(2)e^(2x+3)

From here, I don't know what to do. Please Help.
 
Physics news on Phys.org
There are the little x2 above the typing box. When you have two functions multiplied, you need to use the Product rule to take the derivative.
 
In your differentiation of the particular solution you are forgetting the term that comes from differentiating the ##x^2##.
 
  • Like
Likes   Reactions: scottdave

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
12
Views
2K
Replies
9
Views
2K