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kirbykirby

  • 2 years ago

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

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  1. kirbykirby
    • 2 years ago
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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

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

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

  4. kirbykirby
    • 2 years ago
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    yes

  5. nubeer
    • 2 years ago
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    have you tried Cautiy Eiller..?

  6. kirbykirby
    • 2 years ago
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    WHat's Cautiy Eiller?

  7. chemENGINEER
    • 2 years ago
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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).

  8. kirbykirby
    • 2 years ago
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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|)

  9. nubeer
    • 2 years ago
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    mm no.. sorry that won't work here..

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

  11. kirbykirby
    • 2 years ago
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    after multiplying te integration factor

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

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

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

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

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

  17. chemENGINEER
    • 2 years ago
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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!

  18. kirbykirby
    • 2 years ago
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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?

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

  20. kirbykirby
    • 2 years ago
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    No I was mainly concerned about the absolute value

  21. kirbykirby
    • 2 years ago
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    Thanks though !

  22. shubhamsrg
    • 2 years ago
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    Ignore the absolute value.

  23. UnkleRhaukus
    • 2 years ago
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    yeah.

  24. SWAG
    • 2 years ago
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    Yup

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