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How to solve this differential equation analytically

  1. Jan 8, 2009 #1
    1. The problem statement, all variables and given/known data

    u''(x) - (x^6 + 3*x^2)*u(x) = 0, u(-1) = u(1) = 1, -1 <= x <= 1


    3. The attempt at a solution
    Consider v''(x) - s^2 * v(x) = 0, the auxiliary equation is: r^2 - s^2 = 0. Since (x^6 + 3*x^2) >= 0, s >= 0. Thus, I got r = s or -s. Suppose v(x) = a*exp(s*x)+b*exp(-s*x), then v'(x)=a*s*exp(s*x)-b*s*exp(-s*x), v''(x)=a*s^2*exp(s*x)+b*s^2*exp(-s*x)=s^2*v(x).
    Let s = s(x), u''(x) - s(x)^2 * u(x) = 0, similarly I got u(x) = a*exp(s(x)*x)+b*exp(-s(x)*x). s(x) = Sqrt(x^6 + 3*x^2), u'(x) = a*(Sqrt(x^6 + 3*x^2) + x*(6*x^5+6*x)/(2*Sqrt(x^6 + 3*x^2))) * exp(s(x)*x) + b * (-Sqrt(x^6 + 3*x^2) - x * (6*x^5 + 6*x)/(2*Sqrt(x^6 + 3*x^2))) * exp(-s(x)*x), absolutely, u''(x) will be different from (x^6 + 3*x^2)*u(x), that is to say the method I used is totally incorrect!

    Could anybody tell me what's wrong with my solution and how should I solve this differential equation?

    Thanks a lot!
     
  2. jcsd
  3. Jan 8, 2009 #2
    Your method is incorrect, because there is no auxiliary in this case, since the coefficients are functions and not constants. i.e. you dont have a solution of the form [tex]e^{kx}[/tex]

    For function coefficient second order linear differential there is no general method, to find the solution.
    But if you somehow guess one of the solutions, then an other linearly independent can be formed.

    So in this case a good guess is, search in the form of [tex]e^{f(x)}[/tex]
    Where [tex]f(x)[/tex] is some function, we need to find out.
    plugging this into the equation we get a differential equation for f(x):
    [tex]f''(x)+f'^2(x)=x^6+3x^2[/tex]
    Now we do another trick. Notice:
    [tex]f''=\frac{df'}{df}\frac{df}{dx}=\frac{df'}{df}f'=\frac12 \frac{df'^2}{df}[/tex]

    now put [tex]v=f'^2[/tex].

    So then the DE for f:
    [tex]\frac{1}{2}\frac{dv}{df}+v=x^6+3x^2[/tex]

    this can be solved straight forward, for the [tex]v(f)[/tex] function since it is linear non-homogenous with constant coefficients, and then using [tex]v=f'^2[/tex] we can simply get the [tex]f(x)[/tex] function.
    So we have obtained one solution of the original second order DE, and then using the usual method, you can find a second solution, and then adjust to the boundary conditions and you are done.

    I hope everything was clear :D
     
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