- #1
cefarix
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How would I solve the differential equation y' = - y / ( (x^2 + y^2)^1.5 ) ? Btw, its not a homework prob... thanks
can we replace y' by tan theta?
cefarix said:Btw, its not a homework prob... thanks
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]
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.
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.
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)?
If you're not satisfied with that, do the polar coordinate substitution of the dependent variables.cefarix said:I'm not sure if dy/dx would be the correct form, because then I wouldn't be able to find out the time-dependent equation.