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bahrom7893

  • 5 years ago

Differential equations help! y*dx + (3+3x-y)*dy = 0 The topic is integrating factors The answer is 4xy^3 + 4y^3 - y^4 = C

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  1. anonymous
    • 5 years ago
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    hi bahrom

  2. anonymous
    • 5 years ago
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    want me to use that list thingy

  3. bahrom7893
    • 5 years ago
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    u dont have to

  4. bahrom7893
    • 5 years ago
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    just a solution, its integrating factors, its just my brains are dead

  5. anonymous
    • 5 years ago
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    what do you mean by integrating factors?

  6. bahrom7893
    • 5 years ago
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    you know when you let something be [e^Integral(f(x;y))dx] then multiply across by it to get a derivative of some expression on the right and some ez integral on the left or vice versa

  7. anonymous
    • 5 years ago
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    oh

  8. anonymous
    • 5 years ago
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    so you have y + dy/dx ( 3 + 3x -y) = 0

  9. bahrom7893
    • 5 years ago
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    well yeah then move y over

  10. sasogeek
    • 5 years ago
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    does it mean you're basically trying to prove the answer?

  11. bahrom7893
    • 5 years ago
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    no I just know the answer

  12. anonymous
    • 5 years ago
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    well for integrating factors we want dy/dx + p(x)y = g(x)

  13. anonymous
    • 5 years ago
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    http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

  14. bahrom7893
    • 5 years ago
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    oh okay then i can solve it i think

  15. bahrom7893
    • 5 years ago
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    integrating factor is e^x, lemme try this will let u guys know how it goes

  16. bahrom7893
    • 5 years ago
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    wait no its not, but at least now i know the expression, tryin to solve afk

  17. anonymous
    • 5 years ago
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    here, this is called exact equation http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx

  18. anonymous
    • 5 years ago
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    M(x,y) + N(x,y)dy/dx = 0

  19. bahrom7893
    • 5 years ago
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    k

  20. anonymous
    • 5 years ago
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    dont kill yourself bahrom this website is buggy, everytime i type something it freezes

  21. anonymous
    • 5 years ago
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    still there?

  22. anonymous
    • 5 years ago
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    have you done partial derivatives?

  23. bahrom7893
    • 5 years ago
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    i have, but as i said my brains are not functionin at all, just too tired

  24. anonymous
    • 5 years ago
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    we are looking for a function Fx (x,y) + Fy (x,y) dy/dx = 0, where Fx is the partial derivative of F(x,y) with respect to x (treat y as a constant) and Fy is the partial derivative of F(x,y) with respect to y (treat x as a constant)

  25. anonymous
    • 5 years ago
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    ok .... :(

  26. anonymous
    • 5 years ago
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    We have the equation M(x,y) dx + N(x,y) dy = 0 or M (x,y) + N(x,y)dy/dx = 0 M(x,y) = y and N(x,y) = 3 + 3x - y remember we can solve this and its called "exact" if M(x,y) = Fx (x,y) and N(x,y) = Fy (x,y), where Fx is the partial derivative of F(x,y) with respect to x (treat y as a constant) and Fy is the partial derivative of F(x,y) with respect to y (treat x as a constant) so if we have Fx (x,Y) + Fy (x,y) dy/dx = 0, then F(x,y) is our solution . ok so the last thing , assuming there exists such a F(x,y) it must be the case that Fxy = Fyx which is true for all continuous F(x,y). By substituting Fx = M and Fy = N we have My = Nx . So here Mx = 0 , and Ny = -1 . hmmm, not exact ok now we move to plan B and introduce an integrating factor http://www.cliffsnotes.com/study_guide/Integrating-Factors.topicArticleId-19736,articleId-19711.html

  27. anonymous
    • 5 years ago
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    here is another website http://www.sosmath.com/diffeq/first/exact/exact.html and they have a non exact first order ode

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