
#1
May1208, 04:43 PM

P: 100

How would you attack this equation?
(1/2)y'' + y^2 = 1 I'm trying to find an approximate solution for the trajectory of a ball acted on by constant gravity acceleration and quadratic drag (proportional to the speed squared), and it turns out the speed squared satisfies an equation that is equivalent to this one after nondimensionalizing. Any kind of solution would be nice, whether it is good for large x, small x, or whatever. I was thinking that for x << 1, the equation would be approximately (1/2)y'' + y^2 = 0, in which case y = 3/x^2 is a solution. But then I don't know how to proceed from there to a better approximation. For x >> 1, I was thinking y^2 = 1 would be a first approximation, but again I don't know how to get any farther than that. thanks 



#2
May1208, 04:46 PM

P: 532

I'd multiply through by y', then note that you can integrate each of the three terms individually to get a 1st order diff eqn.
Maybe you probably meant (y')^{2}, in which case try writing as y'' / ( 1  (y')^{2})=2 and integrate. 



#3
May1208, 05:01 PM

P: 100

[tex]\int\frac{dy}{\sqrt{c+y\frac{y^3}{3}}}[/tex] Is there a way to integrate that? thanks 



#4
May1208, 05:10 PM

P: 532

(1/2)y'' + y^2 = 1 



#5
May1208, 05:14 PM

P: 532




#6
May1208, 05:20 PM

P: 100

Also, how do you write an inline fraction, such as 1/2, so that the 1 is on top of the 2? 



#8
May1208, 08:43 PM

P: 100

Well darn. I went back and checked my derivation of the d.e. and realized I had a minus sign wrong. The correct equation is (I pretty sure this time ),
[tex]\frac{1}{2}y'' + 2y^{\frac{1}{2}}y' + y^2 = 1[/tex] So I guess I'll get to work on this one... 



#9
May1308, 02:10 PM

P: 83

Suppose a solution y(x) is invertible locally, then we can define [tex]v(y)=y'(x(y))[/tex], where x(y) is the inverse of y(x). Substituting this in your d.e., you get
[tex]\frac{1}{2} v \dot{v} + 2 y^{1/2} v + y^2 = 1,[/tex] where [tex]\dot{v}[/tex] denotes [tex]dv/dy[/tex]. This is a first order d.e. which is easier to solve. Try substituting [tex]w=v/\sqrt{y}[/tex] to get rid of the square root. If you have found the solution v(y), you can find y(x) by integrating [tex]xx_0 = \int_{y_0}^y \frac{dy}{v(y)}[/tex] and then solving the equation for y. Good luck! ......or use Mathematica . It gives two solution: [tex]y=\tan^2 (xx_0)[/tex] and [tex]y=\coth^2(xx_0)[/tex]. But you should check (and find the general solution). 



#10
May1308, 04:46 PM

P: 100

The problem is, I don't know how to find a general solution for a nonlinear d.e. Is there even a method of knowing how many solutions such an equation has? Perhaps it would be better to work directly with the d.e. for the nondimensionalized velocity, u? u(u'' + 2uu') + (u' + u^{2})^{2} = 1 edit: .............................................. this eqn. comes from the pair of equations u(u'' + 2uu') = sin^{2}A (u' + u^{2})^{2} = cos^{2}A where A(x) is the angle of the velocity vector with the vertical, as a function of the nondimensionalized time x. ................................................ That way the special case of (u' + u^{2})^{2} = 1, which implies u'' + 2uu' = 0, would be apparent. But I'm guessing there are more solutions than that. I will try out your suggestion. Thanks :) 


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