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anonymous

  • 5 years ago

dy/dx=y-2y^2

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  1. anonymous
    • 5 years ago
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    and what do you want?

  2. amistre64
    • 5 years ago
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    ode :)

  3. anonymous
    • 5 years ago
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    solution equation, initial condition is (0,100) i only need a hint on finding:\[\int\limits_{}^{}1\div(y-2y ^{2})\] i dont want much more help beyond that

  4. anonymous
    • 5 years ago
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    I'm not so into the integrals, sorry.

  5. amistre64
    • 5 years ago
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    y' = y -2y^2 it looks like youd just reintegrate each term seperately

  6. anonymous
    • 5 years ago
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    that was the starting equation, i then wrote it as dy/dx and then dy/y-2y^2=dx

  7. anonymous
    • 5 years ago
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    i cant figure out how to take the integral of 1/(y-2yy)

  8. amistre64
    • 5 years ago
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    helps if I keep track of the problem I spose :)

  9. amistre64
    • 5 years ago
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    wiat, which equation you wanting to suit up?

  10. anonymous
    • 5 years ago
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    I think you should use partial fractions

  11. anonymous
    • 5 years ago
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    oh, ok y'=y-2y^2 dy/dx=y-2y^2 dy=(y-2y^2)dx dy/(y-2y^2)=dx integrate both

  12. anonymous
    • 5 years ago
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    ahhh partial fractions!!! that should do it! thanks!!

  13. anonymous
    • 5 years ago
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    aww yeah

  14. anonymous
    • 5 years ago
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    that gives me an imaginary constant of integration -.-

  15. anonymous
    • 5 years ago
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    lol, did the integral work out though? I didn't really read the problem

  16. amistre64
    • 5 years ago
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    would y=e^x ??

  17. amistre64
    • 5 years ago
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    and by that I mean: y = e^x + e^2x +C ?? just wondering

  18. amistre64
    • 5 years ago
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    e^x - e^x^2....

  19. anonymous
    • 5 years ago
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    hold on, ill upload a picture in a sec

  20. anonymous
    • 5 years ago
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    can you just injtegrate the original equation (y'=y-2y^2) to (f(y)+c=1/2 y^2 - 2/3 y^3)

  21. anonymous
    • 5 years ago
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    ?

  22. anonymous
    • 5 years ago
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    I don't see why not, and then use you initial condition for c.

  23. anonymous
    • 5 years ago
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    uilfha thats pathetic, why bother doing the whole dy/dx sub when i could do that ><

  24. anonymous
    • 5 years ago
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    And then I guess solve 0=1/2(100)^2-2/3(100)^3+c for c?

  25. anonymous
    • 5 years ago
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    thats like -661,666.66...

  26. anonymous
    • 5 years ago
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    lol, that whats confusing me. But that seems to be what they mean by the initial condition (0,100) , and thats how you would do it if you wanted the equation for x(y)

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