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kirbykirby

x^2*y' + 3xy = 1 (Linear DE)

  • one year ago
  • one year ago

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  1. kirbykirby
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    I divide by x^2 to get y' + (3/x)y=1/x^2 My integration factor though gives me |x|^3

    • one year ago
  2. kirbykirby
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    I dunno what to do with this absolute value :S

    • one year ago
  3. shamim
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    is it \[x ^{2}y \prime +3xy=1\]

    • one year ago
  4. kirbykirby
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    yes

    • one year ago
  5. nubeer
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    have you tried Cautiy Eiller..?

    • one year ago
  6. kirbykirby
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    WHat's Cautiy Eiller?

    • one year ago
  7. chemENGINEER
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    basically Put the differential equation in the correct initial form Find the integrating factor Multiply everything in the differential equation by and verify that the left side becomes the product rule and write it as such. Integrate both sides, make sure you properly deal with the constant of integration. Solve for the solution y(t).

    • one year ago
  8. kirbykirby
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    (I tried seeing what wolfram alpha does, but they always do Integral (1/x) as ln x, but we are supposed to do it as ln |x|)

    • one year ago
  9. nubeer
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    mm no.. sorry that won't work here..

    • one year ago
  10. kirbykirby
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    I get \[|x|^3y' +\frac{3|x|^3}{x}=\frac{|x|^3}{x^2}\]

    • one year ago
  11. kirbykirby
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    after multiplying te integration factor

    • one year ago
  12. chemENGINEER
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    the final step is then d/dx[x^3y]=int(x) which is simple.

    • one year ago
  13. chemENGINEER
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    excuse me it is x^3y=int(x) which is to say d/dx[x^3y]=x^3/x^2

    • one year ago
  14. kirbykirby
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    But we hae absolute values no? Integral (1/x) = ln|x|

    • one year ago
  15. chemENGINEER
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    this yields x^3y=x^2/2 then jus divide by x^3 to get y=1/2(x^-1)

    • one year ago
  16. kirbykirby
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    How do you know if the x is always positive?

    • one year ago
  17. chemENGINEER
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    Try reading some outside resources like http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx !!! I liked this site when i was a rook!

    • one year ago
  18. kirbykirby
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    COuld it be because if we say x>0, then the drop the absolute value, but if x <0, then you put -x^3 everywhere (but since the whole equation is negative, you just "divide by -1" o_o?

    • one year ago
  19. chemENGINEER
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    otherwise though this solution looks correct..any problems you see?

    • one year ago
  20. kirbykirby
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    No I was mainly concerned about the absolute value

    • one year ago
  21. kirbykirby
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    Thanks though !

    • one year ago
  22. shubhamsrg
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    Ignore the absolute value.

    • one year ago
  23. UnkleRhaukus
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    yeah.

    • one year ago
  24. SWAG
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    Yup

    • one year ago
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