## RolyPoly Group Title Solve by undetermined coefficients/variation of parameters: xy' + y = 1/x (x>0) one year ago one year ago

1. RolyPoly Group Title

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..)

2. RolyPoly Group Title

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

3. RolyPoly Group Title

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

4. nubeer Group Title

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

5. RolyPoly Group Title

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

6. nubeer Group Title

yes i think so.

7. RolyPoly Group Title

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..

8. nubeer Group Title

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

9. nubeer Group Title

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.

10. RolyPoly Group Title

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

11. RolyPoly Group Title

And wolf gives the same answer :(

12. sirm3d Group Title

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$

13. sirm3d Group Title

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

14. RolyPoly Group Title

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

15. sirm3d Group Title

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

16. RolyPoly Group Title

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

17. sirm3d Group Title

^^

18. RolyPoly Group Title

$y_p=u_1y_1+u_2y_2$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?

19. sirm3d Group Title

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

20. RolyPoly Group Title

Yes yes.

21. RolyPoly Group Title

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$$ ?!