Method of Variation of Parameters

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The discussion focuses on solving the differential equation y'' - 2y' + y = e^x/(1+x^2) using the method of variation of parameters. The characteristic equation reveals a repeated root of -1, leading to the homogeneous solution y_h = C_1e^{-x} + C_2xe^{-x}. The Wronskian is calculated as W = e^{-2x}, which is then used to find the functions u_1 and u_2. The user struggles with integrating u_1, specifically the integral of -xe^{2x}/(1+x^2), and realizes a sign mistake in their Wronskian calculation. The conversation highlights common pitfalls in solving differential equations and emphasizes careful verification of calculations.
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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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