Solving a differential equation

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Homework Help Overview

The discussion revolves around solving a differential equation of the form (1+bx)y''(x)-ay(x)=0, which involves variable coefficients. Participants are exploring methods to approach this type of problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply methods for homogeneous linear ordinary differential equations (ODEs) but questions how the presence of the function (1+bx) alters the approach. Some participants suggest dividing both sides by (1+bx) and consider the implications of forming a characteristic equation. Others express uncertainty about the validity of using constant coefficient methods and propose the possibility of clever substitutions or series solutions.

Discussion Status

The discussion is active with participants sharing their thoughts on potential methods and questioning the appropriateness of various approaches. There is a recognition that traditional methods may not apply due to the variable coefficients, and some guidance is offered regarding the use of series solutions.

Contextual Notes

Participants are considering the origin of the differential equation and whether it is derived from a text that may provide additional context or hints for solving it.

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Homework Statement


Solve
(1+bx)y''(x)-ay(x)=0

Homework Equations


The Attempt at a Solution



I'm used to solving homogeneous linear ODE's where you form a characteristic equation and solve that way, here there is the function of x (1+bx) so how does that change things?
 
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Would dividing both sides by 1+bx help?
 
Ok so if I did that then what? I can define a characteristic equation such that

r^2-\frac{a}{1+bx}=0

and r=\pm\sqrt{\frac{a}{1+bx}}

where b^2-4ac = 4a(1+bx) > 0

so a solution is y=ce^{rx} but that doesn't satisfy the ODE so its not correct?
 
You can't use constant coefficient methods on a DE like this with variable coefficients. Perhaps there is a clever substitution that will help, or maybe not. Problems like this are typically solved with series solutions, especially if you know ##a## and ##b##. Where did this equation come from? If it's from a text, the recent material may give a hint how to solve it.
 

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