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RolyPoly

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

  • one year ago
  • one year ago

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  1. RolyPoly
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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..)

    • one year ago
  2. RolyPoly
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    For the left, it looks like Cauchy-Euler equation...

    • one year ago
  3. RolyPoly
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    \[xy_c' + y_c=0\]\[y_c = c_1x^{-1}\] But what about the particular solution?

    • one year ago
  4. nubeer
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    variation of parameter. the other method would not work as we dont have the general for for 1/x

    • one year ago
  5. RolyPoly
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    Are the two equations:\[u_1'(x^{-1}) = 0\]\[u_1'(-\frac{1}{x^2})=\frac{1}{x}\]

    • one year ago
  6. nubeer
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    yes i think so.

    • one year ago
  7. RolyPoly
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    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..

    • one year ago
  8. nubeer
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    hmm i dont know.. but why doesn't look right to you?

    • one year ago
  9. nubeer
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    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.

    • one year ago
  10. RolyPoly
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    The answer to the question is \[y=\frac{c_1}{x}+\frac{logx}{x}\]

    • one year ago
  11. RolyPoly
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    And wolf gives the same answer :(

    • one year ago
  12. sirm3d
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    letting \[y=v(\frac{1}{x})\]\[y'=-\frac{v}{x^2}+\frac{v'}{x}\]\[x\left(-v/x^2 + v'/x\right)+v/x=1/x\]\[v'=\frac{1}{x}\]\[v=\ln x\]

    • one year ago
  13. sirm3d
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    \[y=v\frac{1}{x}=\frac{\ln x}{x}\] the second term of the solution given by wolf @RolyPoly

    • one year ago
  14. RolyPoly
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    How do you come up with y = v/x ???

    • one year ago
  15. sirm3d
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    variation of parameter uses \(y=v\cdot u_1\) where \(u_1\) is a solution to the homogeneous DE. i used \[u_1 = \frac{1}{x}\]

    • one year ago
  16. RolyPoly
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    I... just realized that I didn't (and don't) know variation of parameters works!!! :'(

    • one year ago
  17. sirm3d
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    ^^

    • one year ago
  18. RolyPoly
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    \[y_p=u_1y_1+u_2y_2\]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?

    • one year ago
  19. sirm3d
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    the complementary solution is \[y_c=c_1(\frac{1}{x})\]

    • one year ago
  20. RolyPoly
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    Yes yes.

    • one year ago
  21. RolyPoly
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    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 \(u_2\) ?!

    • one year ago
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