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Frustrating Bernoulli Equation

  1. Feb 16, 2012 #1
    I've been unable to fully solve this: [itex]\frac{dy}{dx} + y = xy^4[/itex]

    The U: [itex]u = y^{-3}[/itex], so [itex]y = u^\frac{-1}{3}[/itex], and [itex]\frac{dy}{dx} = \frac{-1}{3}u^\frac{-4}{3}\frac{du}{dx} [/itex]

    The substitution: [itex]\frac{-1}{3}u^\frac{-4}{3}\frac{du}{dx} + u^\frac{-1}{3} = xu^\frac{-4}{3}[/itex]

    Simplified: [itex]\frac{du}{dx} - 3u = -3x[/itex]

    AKA: [itex]\frac{du}{dx} + 3x = 3u[/itex]

    I've tried solving the resulting equation as a linear:

    [itex]p(x) = 3x[/itex], so the integrating factor is [itex]e^{\frac{3}{2}x^2}[/itex].

    Which creates this unworkable equation: [itex]e^{\frac{3}{2}x^2}\frac{du}{dx} + 3xe^{\frac{3}{2}x^2} = 3ue^{\frac{3}{2}x^2}[/itex]

    And I've tried solving it as a homogenous ([itex]3udx - 3xdx - du = 0[/itex]):

    [itex]u = vx[/itex], so [itex]du = vdx + xdv[/itex]

    Subing those in creates this unworkable mess: [itex]3vxdx - 3xdx - vdx - xdv = 0[/itex]

    Which leaves me stuck, because I've only learned to solve separable, exact, homogenous, linear and bernoullis.
     
  2. jcsd
  3. Feb 16, 2012 #2
    The equation is most easily solved at this stage: [itex]\frac{du}{dx} - 3u = -3x[/itex]

    The integrating factor is just [itex]e^{-3x}[/itex]. Solve from there.
     
  4. Feb 16, 2012 #3

    AlephZero

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    You don't need an integrating factor here. This is a linear equation with constant coefficients. Find the general solution of
    [itex]\frac{du}{dx} - 3u = 0[/itex]
    plus a particular integral of
    [itex]\frac{du}{dx} - 3u = -3x[/itex]
    which is going to be of the form u(x) = a + bx, for some constants of a and b.
     
  5. Feb 16, 2012 #4
    Both methods solve the equation in about 2 lines, but yeah you can do complimentary function and particular integral if you prefer.
     
  6. Feb 16, 2012 #5
    Also the particular integral is u(x) = a - 3x. You don't get two arbitrary constants from a first order ODE
     
  7. Feb 16, 2012 #6
    Sorry it's a + x
     
  8. Feb 16, 2012 #7

    AlephZero

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    You don't get any arbitary constants in a particular integral. My a and b were values to be found, not arbitrary constants.

    No, see above. N.B. I was trying not to give the OP the complete answer to something that might be a homework question.
     
  9. Feb 16, 2012 #8
    Thanks for the help!

    A final question: I thought all integrating factors must have the p(x) part integrated, but this seems to be an exception. How come?
     
  10. Feb 16, 2012 #9

    AlephZero

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    The way I did it is a standard method, but maybe you haven't learned it yet. This logic behind it is:
    For a linear equation like this one, any solution of ##du/dx -3u = -3x## is the sum of the solutions of two equations, the "given" equation
    ##du/dx -3u = -3x##
    and the simpler equation
    ##du/dx - 3u = 0##
    You can easily find the general solution of the equation with 0 on the right hand side, and that solution ##Ce^{3x}## contans an arbitrary constant.

    You know there is only one arbitrary constant in the complete solution, so you don't need another one. Instead of finding the general solution of ##du/dx -3u = -3x##, now we only need one particular solution. You can usually guess the form of a particular solution from the function on the right hand side, and then equate the coefficients of different terms in the equation to find the unknown values.

    In this case, we can guess that ##u = ax + b## is a solution for some values of a and b. Then ##du/dx = a##, and the substituting in the differential equation we have
    ##a -3(ax + b) = -3x##
    To satisfy that equation for all values of ##x##, the constant terms and the coefficients of ##x## must both cancel out. So we have
    ##a - 3b = 0##
    and
    ##-3ax = -3x## for all ##x##
    which gives ##a = 1## and ##b = 1/3##
    and the general solution is ##u = Ce^{3x} + x + 1/3##.
    That method might seem longwinded (or even sneaky) but the real benefit comes because most of the "interesting" differential equations in physics and engineering are second order not first order, and finding integrating factors for second order equations is hard. (And most of the longwindedness was explaining it, not actually doing it).

    Alternatively you can solve this equation using an integrating factor without any "trickery".
    ##du/dx -3u = -3x##
    The integrating factor is ##e^{-3x}##
    ##e^{-3x}du/dx - 3u e^{-3x} = -3xe^{-3x}##
    ##d/dx (ue^{-3x}) = -3xe^{-3x}##
    ##ue^-3x = -3\int x e^{-3x}\,dx##
    And you need to integrate by parts to do the integral on the right hand side.
     
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