Method of Variation of Parameters

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Homework Help Overview

The problem involves solving a second-order linear differential equation using the method of variation of parameters. The equation presented is y'' - 2y' + y = e^x / (1 + x^2), with a focus on finding a particular solution after determining the homogeneous solution.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the Wronskian and its implications for the method of variation of parameters. There are attempts to clarify the roots of the characteristic equation and the integration process involved in finding the particular solution.

Discussion Status

The discussion is ongoing, with participants identifying potential errors in calculations and clarifying the correct roots of the characteristic equation. Some guidance has been offered regarding the Wronskian and integration challenges, but no consensus has been reached on the correct approach to the integration problem.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is an acknowledgment of mistakes in calculations that may affect the overall understanding of the problem.

ElijahRockers
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Homework Statement



y''-2y'+y = \frac{e^x}{1+x^2}

Homework Equations



u_1 = -\int \frac{y_{2}g(x)}{W}dx
u_2 = \int \frac{y_{1}g(x)}{W}dx
g(x) = \frac{e^x}{1+x^2}
W is the wronskian of y1 and y2.

The Attempt at a Solution



The characteristic equation for the homogenous solution yields a repeated root of -1, so

y_{h} = C_{1}e^{-x} + C_{2}xe^{-x}

When I calculated the Wronskian it simplified to

W = e^{-2x}

Plugging in the formula for u1 and simplifying I get

u_{1} = -\int \frac{xe^{2x}}{1+x^2}dx

I'm not quite sure how to go about solving this. My best shot at integration by parts didn't really seem to help. I ended up having to integrate ln(x^2 + 1)e^2x which doesn't seem any easier.
 
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I got W=(1-2x)e-2x.
 
Hmmm, alright.. I can't seem to find my mistake

if W is
|y1 y2|
|y'1 y'2|

I got
y1 = e^-x
y2 = xe^-x
y'1 = -e^-x
y'2 = -xe^-x + e^-x = e^-x(1-x) (after factoring off the e^-x)

and y1y'2 - y'1y2 = e^(-2x)(1-x) + xe^(-2x) = e^(-2x)
 
Sorry, I made a sign mistake.
 
Oh, your roots are wrong. They should be +1.
 
O geez... always something little.

Thanks!
 

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