Solve Euler's Equation: x3y'''+15x2y''+61xy'+64y=0

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Discussion Overview

The discussion revolves around solving the homogeneous differential equation: x3y''' + 15x2y'' + 61xy' + 64y = 0. Participants explore the nature of the solutions and the implications of their findings, focusing on the methods of solving differential equations and the application of the superposition principle.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant claims to have found three solutions, all equal to -4, and proposes a general solution of the form y = c1x-4 + c2x-4ln(x) + c3x-4ln(2x).
  • Another participant expresses agreement with the proposed solution.
  • A third participant notes that if the proposed solutions are valid, then by the superposition principle, their sum would also be a solution to the differential equation.
  • A later reply discusses a substitution t = ln(x), transforming the original equation into one with constant coefficients, leading to a characteristic equation and a general solution that aligns with the earlier proposed solution.

Areas of Agreement / Disagreement

Participants generally agree on the form of the solution and the application of the superposition principle, but the discussion does not resolve whether the initial claims about the solutions are definitively correct.

Contextual Notes

The discussion includes transformations and substitutions that may depend on specific assumptions about the nature of the solutions and the domain of x. The implications of these transformations are not fully explored.

manenbu
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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
?
 
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I think you get it right. :approve:
 
If they are all solutions to the differential equation, then, by the superposition principle, the sum is a solution.
 
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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