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Main Question or Discussion Point
How would I solve the differential equation y' = - y / ( (x^2 + y^2)^1.5 ) ? Btw, its not a homework prob... thanks
It's not quite that simple. First, you need to realize that everything is now a function of θ, so you're going to have to do things like the chain rule.can we replace y' by tan theta?
The idea behind moving homework questions to the Homework section is to clear up the Math section for discussion of mathematical topics, without having this kind of step-by-step help discussions hanging around. While this may not be a homework problem that was assigned to you, it looks just like one, and it is handled just like one, so it really should be posted there.cefarix said:Btw, its not a homework prob... thanks
Should be:cefarix said:[tex]\frac{dy}{dx} = - \frac{r \cos \theta + \frac{dr}{d\theta} \sin \theta}{r \sin \theta + \frac{dr}{d\theta} \cos \theta}[/tex]
You know what? I think I made a mistake moving this thread--sorry. When I initially looked at it I assumed that it would have a quick, closed-form solution. But it's clear now that that is not the case, and that this could lead to an interesting mathematical discussion. So, I'm sending it to Differential Equations.Tom Mattson said:The idea behind moving homework questions to the Homework section is to clear up the Math section for discussion of mathematical topics, without having this kind of step-by-step help discussions hanging around. While this may not be a homework problem that was assigned to you, it looks just like one, and it is handled just like one, so it really should be posted there.
Thanks Hurkyl. That's an interesting approach. Seems though the actual asymptote cannot be determined analytically without the analytical solution in hand.Hurkyl said:It doesn't look like the x-axis is the asymptote for those plots.
I don't really know if this approach is kosher, but...
I made the change of variable:
y → y(t) / z(t)
x → x(t) / w(t)
from which I get the differential equation:
[tex]
(y' z - y z') (x^2 z^2+ y^2 w^2)^{3/2} = (x'w - xw') z^4 y w
[/tex]
where differentiation is with respect to t.
Now, notice that when w(a) = 0 (corresponding to infinite x in the original problem), I can satisfy the differential equation
y'z - yz' = 0
Or equivalently,
y = Kz
meaning that the original y(t) is roughly the constant K when t is near a.
Not sure what you mean. I posted three particular solutions to the first equation above (numerically determined but particular nevertheless). What's with the x'? With respect to y or a coupled sytem with respect to t?cefarix said:Can someone show me a particular solution of the two equations y' = -y/((x^2+y^2)^1.5) and x' = -x/((x^2+y^2)^1.5)?