Homogeneous linear ODEs with Constant Coefficients

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Discussion Overview

The discussion revolves around homogeneous linear ordinary differential equations (ODEs) with constant coefficients, focusing on the nature of their solutions and specific examples. Participants explore different types of solutions and their validity under various conditions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the equation $ay' + by = 0$ and derives that $y' = -\frac{b}{a}y$, suggesting that the solutions are exponential functions.
  • Another participant counters by proposing that $y = \sin x$ is also a valid solution for the equation $y'' + y = 0$.
  • A subsequent reply reiterates the validity of $y = \sin x$ and mentions that it is a solution as long as certain boundary conditions are not imposed.
  • The same participant also notes that $y = A\sin x + B\cos x$ can be expressed in terms of an exponential function, indicating a connection between trigonometric and exponential solutions.

Areas of Agreement / Disagreement

Participants express differing views on the types of functions that can serve as solutions to the discussed ODEs. There is no consensus on the exclusivity of exponential functions as solutions, as trigonometric functions are also proposed.

Contextual Notes

The discussion does not resolve the implications of boundary conditions on the solutions presented, nor does it clarify the conditions under which each type of solution is valid.

oasi
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do you have a idea about it?can you help me

http://img17.imageshack.us/img17/1156/18176658.png
 
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oasi said:
do you have a idea about it?can you help me

http://img17.imageshack.us/img17/1156/18176658.png

For example, take $ay' + by = 0$. Solving for y' yields
$$
y' = -\frac{b}{a}y = ky
$$
where k = -b/a.

The only nontrivial function whose derivative is a constant multiple of itself is an exponential function.
 
But $y = \sin x$ could work. For example,

$y'' + y = 0$

has as one solution $y = \sin x$.
 
Jester said:
But $y = \sin x$ could work. For example,

$y'' + y = 0$

has as one solution $y = \sin x$.

As long as the boundaries aren't $y'(0) = 0$ and $y'\left(\frac{\pi}{2}\right) = 0$ then y = 0. :)

But $y = A\sin x + B\cos x = e^0\left(A\sin x + B\cos x\right)$ is also a solution of the non boundary value problem.
 

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