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Solve by undetermined coefficients/variation of parameters: xy' + y = 1/x (x>0)

Differential Equations
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I know using integrating factor is a better way (at least for me), but still I would like to do it using these two methods (or at least one of them..)
For the left, it looks like Cauchy-Euler equation...
\[xy_c' + y_c=0\]\[y_c = c_1x^{-1}\] But what about the particular solution?

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Other answers:

variation of parameter. the other method would not work as we dont have the general for for 1/x
Are the two equations:\[u_1'(x^{-1}) = 0\]\[u_1'(-\frac{1}{x^2})=\frac{1}{x}\]
yes i think so.
Hmm... \[u_1'(-\frac{1}{x^2})=\frac{1}{x}\]\[u_1' = -x\]\[u_1=-\frac{x^2}{2}\]? But that doesn't seem right..
hmm i dont know.. but why doesn't look right to you?
http://www.wolframalpha.com/input/?i=xy%27+%2B+y+%3D+1%2Fx ok try to look on this.. as it's not opening up for me.
The answer to the question is \[y=\frac{c_1}{x}+\frac{logx}{x}\]
And wolf gives the same answer :(
letting \[y=v(\frac{1}{x})\]\[y'=-\frac{v}{x^2}+\frac{v'}{x}\]\[x\left(-v/x^2 + v'/x\right)+v/x=1/x\]\[v'=\frac{1}{x}\]\[v=\ln x\]
\[y=v\frac{1}{x}=\frac{\ln x}{x}\] the second term of the solution given by wolf @RolyPoly
How do you come up with y = v/x ???
variation of parameter uses \(y=v\cdot u_1\) where \(u_1\) is a solution to the homogeneous DE. i used \[u_1 = \frac{1}{x}\]
I... just realized that I didn't (and don't) know variation of parameters works!!! :'(
^^
\[y_p=u_1y_1+u_2y_2\]In this case, only one complementary solution, so \[y_p = u_1y_1\]where \(y_1=x^{-1}\) \[y_p'=\frac{u_1'}{x}-\frac{u_1}{x^2}\] Sub these into the DE and solve \(u_1\) Is this the idea?
the complementary solution is \[y_c=c_1(\frac{1}{x})\]
Yes yes.
In case there are two, I can use the equations \[u_1'y_1+u_2'y_2 =0\]\[u_1'y_1'+u_2'y_2' =f\]to get \(u_1\) and \(u_2\) ?!

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