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Quick question (Euler)

  1. Jan 2, 2010 #1
    Given the homogeneous equation:
    x3y''' + 15x2y'' + 61xy' + 64y = 0

    I get 3 solutions which are all -4.

    Does this mean that the solution for y should be:

    y = c1x-4 + c2x-4lnx + c3x-4ln2x
  2. jcsd
  3. Jan 2, 2010 #2
    I think you get it right. :approve:
  4. Jan 2, 2010 #3
    If they are all solutions to the differential equation, then, by the superposition principle, the sum is a solution.
  5. Jan 3, 2010 #4


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    Notice, by the way, that the substitution t= ln(x) changes the differential equation into the "constant coefficients" equation [itex]d^3y/dt^3+ 12d^2y/dt^2+ 48dy/dt+ 64y= 0[/itex] which has characteristice equation [itex]r^3+ 12r^2+ 48r+ 64= (r+ 4)^2= 0[/itex] so the general solution to the constant coefficients equation is [itex]y(t)= C_1e^{-4t}+ C_2te^{-4t}+ C_3t^2 e^{-4t}[/itex] and the general solution to the original equation is [itex]y(x)= C_1e^{-4ln(x)}+ C_2 ln(x) e^{-4ln(x)}[/itex][itex]+ C_3 (ln(x))^2 e^{-4ln(x)}[/itex][itex]= C_1x^{-4}[/itex][itex]+ C_2 ln(x) x^{-4}+ C_3 (ln(x))^2 x^{-4}[/itex], just as you say.
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