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Solve differential equation with boundary conditions using substitution

  1. Nov 4, 2012 #1
    μ[itex]^{2}[/itex][itex]\frac{d^{2}u}{dx^{2}}[/itex]+ae[itex]^{u}[/itex]=0

    Boundary conditions: u(-L)=u(L)=u[itex]_{0}[/itex]

    Solve by multiplying by [itex]\frac{du}{dx}[/itex] and integrating in x

    I know you have to use substitution, but I keep going in circles.
     
  2. jcsd
  3. Nov 5, 2012 #2
    The way that I would do it would be to rewrite the equation as
    [itex]\mu[/itex][itex]^2[/itex] [itex]'' = -ae^\mu[/itex]

    Then you solve for a general solution which is
    [itex]\mu(x) = 1 + x[/itex]

    Then you go on to get your general solution from the boundaries and such.

    Another method, which would probably be easier and is what I think you were trying to do is to just integrate twice and get:
    [itex]\frac{\mu^4}{12} = -ae^\mu + \mu*c_1 + c_2[/itex]

    That may be slightly off, but try integrating it yourself, and then use your boundary conditions to solve for the constants.
     
    Last edited: Nov 5, 2012
  4. Nov 5, 2012 #3
    Hi !
    I think that the two methods proposed by danielu13 are both incorrect.
    Better follow the advice given in the wording : multiply by u'
    m² u'' +a exp(u) = 0
    m² 2 u''u' +2a exp(u) u' = 0 and integrate :
    m² u'² +2a exp(u) = c
    m u' = (+ or -)sqrt(c-2a exp(u))
    Case +sqrt :
    m (du/dx) = sqrt(c-2a exp(u))
    dx = m du / sqrt(c-2a exp(u))
    Then integrate, wich gives x(u)
    and invert it to obtain u(x).
     
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