What is the variation of parameter method for solving differential equations?

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Homework Help Overview

The discussion centers around the variation of parameters method for solving a specific second-order linear differential equation, represented as (D^2 + 2D + 1)y = ln(x)/(xe^x). The original poster attempts to find a particular solution and expresses their work through various substitutions and integrations.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster outlines their approach to solving the differential equation, including finding the roots and deriving equations for coefficients. Some participants provide feedback on the correctness of the solution and suggest minor adjustments regarding constants.

Discussion Status

The discussion is ongoing, with participants engaging in verification of the original poster's calculations and providing informal feedback. There is no explicit consensus on the final form of the solution, but some guidance has been offered regarding the constants involved.

Contextual Notes

The original poster expresses a desire to understand the forum's features, indicating they are new to the platform. This context may influence their engagement in the discussion.

jbord39
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Homework Statement



(D^2 + 2D + 1)y = ln(x)/(xe^x)

Homework Equations



D = d/dx

The Attempt at a Solution



First I find the roots of the left side of the equation, -1 of multiplicity 2.
This leads to
y(c) = Ae^(-x) + Bxe^(-x)

Substituting A and B with a' and b' and dividing both sides by e^(-x) I find the two equations:

-a' + (1-x)b' = ln(x)/x
a' + xb' = 0

Which leads to a' = -ln(x) and b' = ln(x)/x

Integrating by parts leads to:

a = x(1-ln(x))
b = (1/2)[ln(x)]^2

Which leads to

y(p) = ae^(-x) + bxe^(-x)
= xe^(-x)[1-ln(x) + (1/2)(ln(x))^2]

So y = xe^(-x)[1 - ln(x) + (1/2)(ln(x))^2 + A/x + B]

Does this seem correct?
 
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It matches what I get. You might notice that the '1' doesn't need to be there, you could just merge it into the constant B.
 
Thanks a bunch for the reply. I am new to these forums and wondering if there is a "Thank" button or anything like I've seen on other similar forums.
 
jbord39 said:
Thanks a bunch for the reply. I am new to these forums and wondering if there is a "Thank" button or anything like I've seen on other similar forums.

Not that I know of. But thanks for looking for it.
 

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