ODE y''+b^2 y=0 where b is vector

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SUMMARY

The discussion centers on solving the ordinary differential equation (ODE) y'' - b²y = 0, specifically when b is a vector represented as \(\vec{b} = (b_x, b_y)\). The proposed solutions include y = C e^{\pm \vec{b}x} and y = C e^{\pm |b| x}. It is established that when b is a vector, "b²" refers to the dot product, leading to the conclusion that the coefficient is the square of the vector's length, |b|², which influences the solution form.

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mogul
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Hi,
please help me with this task. I'm wondering what is the right result.
I have a ODE
[itex]y'' - b^2 y =0[/itex]

also the result should be
[itex]y=C e^{\pm bx}[/itex]

but what is the result when b is vector?
[itex]\vec b=(b_x, b_y)[/itex]

is this the result?
[itex]y=C e^{\pm \vec{b}x}[/itex]

or this?
[itex]y=C e^{\pm |b| x}[/itex]

how to solve it?
thanks for help!
 
Physics news on Phys.org
If b is a vector, then what do you mean by "[itex]b^2[/itex]"? The dot product? Then coefficient is the square of the length of the vector.
 

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