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

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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