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Linear ODE Non-constant coefficient

  1. Jan 19, 2010 #1
    Hi. I'm having difficulty remembering how to solve for u(r).
    The equation is r*u''+u'=0 with BC u(2)=20; u(1)=540.

    Any help would be appreciated. I really need help setting up how to solve. Thanks.
  2. jcsd
  3. Jan 19, 2010 #2
    This is an Euler equation (a standard form would be with u'' coefficient being r^2, so just multiply the equation by r), and is solved with guessing a solution [tex]u(r)=r^{\lambda}[/tex]
    Substituting into the equation gives:

    [tex]\lambda (\lambda-1)r^{\lambda -1}+\lambda r^{\lambda-1}[/tex]

    Then divding by the power of r gives you the characteristic polynomial:

    [tex] P(\lambda)= \lambda ^ {2} =0 [/tex]

    So you have one solution [tex]u(r)=r^{0}=1[/tex], the second solution will e [tex]u(r)=ln(r) r^{0}=ln(r)[/tex] (which is a result of a manipulation analogous to constant coefficient theory)

    And a general solution is [tex]u(r)=A+B ln(r)[/tex], and then just use initial conditions
  4. Jan 19, 2010 #3
    Thanks so much. It reminded me of Cauchy-Euler, but I did not know how to approach it with lambda=0.
  5. Jan 20, 2010 #4


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    Another way to do this problem is to note that u does not appear explicitely in the problem.

    Let v= y' and the equation becomes rv'+ v= 0, a simple, separable, first order equation.

    rv'= -v so dv/v= -dr/r and, integrating, ln v= ln -r+ C or v= C'/r.

    Now we have u'= C'/r so integrating again, u(t)= C'/2 ln r+ C", exactly what elibj123 got.
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