Homogeneous linear ODEs with Constant Coefficients

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SUMMARY

This discussion focuses on homogeneous linear ordinary differential equations (ODEs) with constant coefficients, specifically the equation $ay' + by = 0$. The solution involves finding the derivative $y' = -\frac{b}{a}y = ky$, where $k = -b/a$. The primary solution type is the exponential function, although trigonometric functions like $y = \sin x$ also serve as solutions under certain conditions, such as $y'' + y = 0$. The discussion emphasizes that boundary conditions significantly influence the solution set.

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  • Understanding of homogeneous linear ordinary differential equations
  • Familiarity with constant coefficients in differential equations
  • Knowledge of exponential and trigonometric functions
  • Basic concepts of boundary value problems
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  • Study the method of solving homogeneous linear ODEs with constant coefficients
  • Explore the role of boundary conditions in determining solutions
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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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