Re: differential equation $\ddot x + a \dot x + \frac{ b }{ 1 + c x } + d = 0$

Actually, the first-order equation is from a class called Abel's equation
of the second kind. But just because it has a name does not mean that
anyone knows how to solve it. So numerical methods are probably the
way to go.

--
Daniel Mayost


In article <1162314147.468333.203660@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<isscandar@xxxxxxxxx> wrote:
Daniel Mayost wrote:
I don't think this equation has a special name. As far as solving it
goes, since the independent variable does not appear in it, you can
make the substitution x'=y(x). Then, x''=yy', and the equation becomes:

yy' + ay + b/(1+cx) + d = 0

This is now a first-order equation instead of a second-order equation.
I don't know where to go from here though.

thank you,
I tried to solve analitically but I stuck at this

dy(t)/dt ^ 2 = - 2cy(t) - 2 a / b ln( b / a y(t) + 1 / a )

$\frac{dy(t)}{dt}^2 = - 2cy(t) - 2\frac{a}{b} ln( \frac{b}{a} y(t) +
\frac{1}{a} )$

I'll try numerical solution...


--
Daniel Mayost

Alessandro



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