MHB Homogeneous linear ODEs with Constant Coefficients

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.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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