Recent content by CGZ
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Undergrad How to Solve the Differential Equation (x^2 - 1)y' + 2xy = x?
(x^2 - 1 ) y' + 2xy = ((x^2-1)y)' = x;--> (x^2-1)y = \int{x};--> (x^2-1)y = 0.5 x^2 + C y = (0.5 x^2 + C) / ( x^2 - 1 )- CGZ
- Post #7
- Forum: Differential Equations
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Zeros of bessel functions in scilab
I do not use scilab, but i have a soft ware can do this. Here is the software:http://www.mwtee.com/thread-679058-1-1.html- CGZ
- Post #2
- Forum: MATLAB, Maple, Mathematica, LaTeX