Help finding a general solution for an eigenvalue problem

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SUMMARY

The discussion focuses on solving the second-order homogeneous ordinary differential equation (ODE) given by X''(x) + X'(x) - λX(x) = 0, which is a crucial step in addressing an eigenvalue problem. The characteristic equation m^2 + m - λ = 0 leads to the expression for m involving the term √(1 + 4λ). The solution requires analyzing three cases based on the value of 1 + 4λ: positive, zero, and negative. Participants emphasize the importance of correctly applying the quadratic formula to derive the roots of the characteristic equation.

PREREQUISITES
  • Understanding of second-order homogeneous ordinary differential equations (ODEs)
  • Familiarity with eigenvalue problems in differential equations
  • Knowledge of the quadratic formula and its application
  • Basic concepts of characteristic equations
NEXT STEPS
  • Study the implications of different cases for eigenvalues in ODEs
  • Learn about boundary value problems and their solutions
  • Explore the method of undetermined coefficients for solving ODEs
  • Investigate the role of the discriminant in quadratic equations
USEFUL FOR

Students studying differential equations, mathematicians tackling eigenvalue problems, and educators seeking to clarify the solution process for second-order ODEs.

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Homework Statement


Hey, guys. I'm having trouble finding the general solution to a second order, homogeneous ODE. It is the first step to solving an eigenvalue problem and my professor is about as much help as a hole in the head. I've tried multiple "guesses" and have combed various resources looking for general solutions and have come up empty. Can someone help out, please?

Homework Equations


X''(x) + X'(x) - λX(x) = 0
X(0) = 0, X(a) = 0

The Attempt at a Solution


I cannot even solve the characteristic equation, m^2 + m - λ = 0 because you end up getting √1+4λ under the radical in the quadratic equation. In summary, I'm stuck, haha.
 
Last edited:
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What's wrong with that? [itex]\lambda[/itex] is just a number. There is no reason in the world why it could not be "under the radical".

By the way, it is not [itex]\sqrt{1- \lambda^2}[/itex]. By the quadratic formula, the solution to [itex]m^2+ m- \lambda= 0[/itex] is
[tex]\frac{-1\pm\sqrt{1^2- 4(1)(-\lambda)}}{2(1)}= \frac{-1\pm\sqrt{1+ 4\lambda}}{2}[/tex]

Now, you would want to consider the cases [itex]1+ 4\lambda> 0[/itex], [itex]1+ 4\lambda= 0[/itex], and [itex]1+ 4\lambda< 0[/itex] separately.
 

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