How to Obtain Fundamental Solutions for Non-Constant Coefficient Equations?

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SUMMARY

The discussion focuses on solving non-constant coefficient differential equations using the method of variation of parameters. Specifically, the equation discussed is (x^2)y'' - (4x)y' + 6y = x^4*sinx, where the corresponding homogeneous equation is y'' - (4/x)y' + (6/x^2)y = 0. Participants emphasize the importance of finding fundamental solutions of the form y(x) = x^r, and suggest using the Frobenius method for deriving these solutions. The key takeaway is that substituting y(x) = x^r into the homogeneous equation allows for determining valid values of r.

PREREQUISITES
  • Understanding of differential equations, particularly second-order linear equations.
  • Familiarity with the method of variation of parameters.
  • Knowledge of the Frobenius method for solving differential equations.
  • Basic grasp of the concept of fundamental solutions in the context of differential equations.
NEXT STEPS
  • Study the method of variation of parameters in detail, focusing on non-constant coefficients.
  • Learn about the Frobenius method and its application to differential equations with variable coefficients.
  • Explore the derivation of fundamental solutions for second-order linear differential equations.
  • Practice solving various non-constant coefficient differential equations to reinforce understanding.
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Mathematics students, educators, and professionals dealing with differential equations, particularly those interested in advanced methods for solving non-constant coefficient equations.

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Solve by method of variation of parameters
(x^2)y'' - (4x)y' + 6y = x^4*sinx (x > 0)

Hey, I know how to solve problems using variation of parameters but only when the corresponding homogenous equation has constant coefficients...

y'' - (4/x)y' + (6/x^2)y = 0.. the bit I am confused about is how to obtain the fundamental solutions to this equation {y1, y2} when the coefficients are not constants. Any help would be appreciated.

Thanks.
 
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Note that each term has the same "units", if you think of x as having units of length (so that a derivative removes one unit of length). Such differential equations have solutions of the form x^r. Plug this in and solve for r, and you'll quickly see why such solutions work. This is something you can just remember, although it would also fall out if you tried the Frobenius method.
 
im not quite sure I understand where to plug in x^r .
 
I mean y(x)=xr is a solution to the homogenous differential equation for certain r. Plug in this y and see which r work.
 

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