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Math Help - separable differential equation

  1. #1
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    separable differential equation

    separable differential equation-image.png

    In the above I get y^3+y = x^2 +5x +c

    but then if it passes through (1,0) then when x=1, y =0. That gives me 0=7 so I must have gone wrong somewhere. I'm not sure what I'm doing wrong here.

    thanks

    MH
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  2. #2
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    Re: separable differential equation

    $\dfrac {dy}{dx}=\dfrac{4x+5}{3y^2+1}$

    $(3y^2+1)~dy = (4x+5)~dx$

    $y^3+y = 2x^2 + 5x + C$

    we need to find $C$ to satisfy the constraint that the curve passes through $(1,0)$

    $0=2(1^2) + 5(1) + C$

    $0=2 + 5 + C$

    $C=-7$

    The implicit form of your curve is thus

    $y^3+y=2x^2+5x-7$

    This will have an explicit form for $y$ in terms of $x$ but it will be a mess. The implicit form should suffice.
    Thanks from kingsolomonsgrave
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  3. #3
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    Re: separable differential equation

    OH thanks much!
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