Question

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

Answer 1

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..)

Answer 2

For the left, it looks like Cauchy-Euler equation...

Answer 3

\[xy_c' + y_c=0\]\[y_c = c_1x^{-1}\]
But what about the particular solution?

Answer 4

variation of parameter. the other method would not work as we dont have the general for for 1/x

Answer 5

Are the two equations:\[u_1'(x^{-1}) = 0\]\[u_1'(-\frac{1}{x^2})=\frac{1}{x}\]

Answer 6

yes i think so.

Answer 7

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..

Answer 8

hmm i dont know.. but why doesn't look right to you?

Answer 9

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.

Answer 10

The answer to the question is \[y=\frac{c_1}{x}+\frac{logx}{x}\]

Answer 11

And wolf gives the same answer :(

Answer 12

letting \[y=v(\frac{1}{x})\]\[y'=-\frac{v}{x^2}+\frac{v'}{x}\]\[x\left(-v/x^2 + v'/x\right)+v/x=1/x\]\[v'=\frac{1}{x}\]\[v=\ln x\]

Answer 13

\[y=v\frac{1}{x}=\frac{\ln x}{x}\] the second term of the solution given by wolf @RolyPoly

Answer 14

How do you come up with y = v/x ???

Answer 15

variation of parameter uses \(y=v\cdot u_1\) where \(u_1\) is a solution to the homogeneous DE. i used \[u_1 = \frac{1}{x}\]

Answer 16

I... just realized that I didn't (and don't) know variation of parameters works!!! :'(

Answer 17

^^

Answer 18

\[y_p=u_1y_1+u_2y_2\]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?

Answer 19

the complementary solution is \[y_c=c_1(\frac{1}{x})\]

Answer 20

Yes yes.

Answer 21

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 \(u_2\) ?!