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Math Help - Differential equation in implicit form

  1. #1
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    Differential equation in implicit form

    I was wondering if someone could give me some idea as where to start with the following question?

    Find in implicit form the general solution of the differential equation

    dy/dx\ = -3 y^3 (1 + e^{-3x} )(3x - e^{-3x} + 3)

    Any help would be greatly appreciated

    Thanks
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  2. #2
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    Quote Originally Posted by bobred View Post
    I was wondering if someone could give me some idea as where to start with the following question?

    Find in implicit form the general solution of the differential equation

    dy/dx\ = -3 y^3 (1 + e^{-3x} )(3x - e^{-3x} + 3)

    Any help would be greatly appreciated

    Thanks
    Your equation is separable

    \frac{dy}{y^3} = -3 (1 + e^{-3x} )(3x - e^{-3x} + 3) dx

    expand the RHS and integrate term by term.
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  3. #3
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    Quote Originally Posted by bobred View Post
    I was wondering if someone could give me some idea as where to start with the following question?

    Find in implicit form the general solution of the differential equation

    dy/dx\ = -3 y^3 (1 + e^{-3x} )(3x - e^{-3x} + 3)
    Divide through by -y^3 and integrate: -\int\frac{dy}{y^3} = \int 3(1 + e^{-3x} )(3x - e^{-3x} + 3)dx. The x-integral is easy, because that function is the derivative of \tfrac12(3x - e^{-3x} + 3)^2.
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    Thanks to you both, it's clear now.

    James
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  5. #5
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    Hi

    I take -\int\frac{dy}{y^3} is the same as -\int\frac{1}{y^3}dy in which case I find the explicit form and solve for y=1/2 and x=0.

    But the next part asks for the explicit form which I have had a go at, when I use y=1/2 and x=0 I get a different answer is this correct?

    Thanks
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  6. #6
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    Quote Originally Posted by bobred View Post
    Hi

    I take -\int\frac{dy}{y^3} is the same as -\int\frac{1}{y^3}dy in which case I find the explicit form and solve for y=1/2 and x=0.

    But the next part asks for the explicit form which I have had a go at, when I use y=1/2 and x=0 I get a different answer is this correct?
    When you do the integration you should get the implicit form of the solution as y^{-2} = (3x-e^{-3x}+3)^2 + const. If the initial condition is that y=1/2 when x=0 then the constant is 0. You can then take the square root of both sides to get the explicit solution as y = \pm\frac1{3x-e^{-3x}+3}. Finally, check the initial condition again to see that you need the + sign, not the sign.
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  7. #7
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    Thanks, I'd been looking at it for ages and going round and round in circles.
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  8. #8
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    Quote Originally Posted by Opalg View Post
    When you do the integration you should get the implicit form of the solution as y^{-2} = (3x-e^{-3x}+3)^2 + const. If the initial condition is that y=1/2 when x=0 then the constant is 0. You can then take the square root of both sides to get the explicit solution as y = \pm\frac1{3x-e^{-3x}+3}. Finally, check the initial condition again to see that you need the + sign, not the – sign.
    I might be being totally blonde here, but isn't the square root of y^-2 = y^-1?
    Last edited by Confuzed; March 16th 2009 at 04:07 AM.
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  9. #9
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    Reciprocal

    Quote Originally Posted by Confuzed View Post
    I might be being totally blonde here, but isn't the square root of y^-2 = y^-1?
    Yes you're right, but then take the reciprocal of both sides.
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