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Topic: Solving a simple differential equation
Replies: 18   Last Post: Feb 5, 2007 4:30 AM

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Han de Bruijn

Posts: 5,094
Registered: 12/13/04
Re: Solving a simple differential equation
Posted: Feb 5, 2007 4:30 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply wrote:

> On 5 Feb, 08:22, Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:

>> wrote:

>>>Solve the following differential equation:
>>>Solve for x in terms of t.

>>Are you kidding? This is the same as your "Simple Physics Question".
>>All answers have been given in that thread, mainly by the author who
>>calls himself -Rotwang (

> Thanks for the namecheck, though the truth is that I never gave an
> actual solution to the above equation. In fact the post to which you
> are replying came before the "Simple Physics Question" thread, in
> which the author wrote, "I asked this question before but I think I
> phrased it incorrectly."; in the later thread the OP only asks for the
> time at which x=0, which I got without solving the e.o.m. for x. The
> lesson for the OP if he's still reading is that the most obvious way
> of solving a problem is not always the easiest...

Anyway, your arrival time is correct. But *I* have made another mistake
while attempting to prove it in "Simple Physics Question". This is the
big picture:

t = T <- v t = 0
M --------- R ------------ m
x = 0 x -> x = R

The boundary condition (v = 0) for (t = 0) is guaranteed by:

v^2/2 = GM/x - GM/R ==> dx/dt = - sqrt(2.GM/x - 2.GM/R)

Where a minus sign corrects the error I've made. The right time is:

t = sqrt(R^3/(2.GM)) [sqrt(x/R.(1-x/R)) - arcsin(2.x/R-1)/2 + pi/4]

Where (t = 0) for (x = R) : the other boundary condition.

Only _now_ leading to your correct arrival time (T) for (x = 0).

This has been programmed in Delphi Pascal (following an advice from

for coding the ArcSin function: )

function ArcSin(x : double) : double;
ArcSin := 2 * ArcTan(x/(1+Sqrt(1-x*x)));

procedure TForm1.tekenen(Sender: TObject);
W,H,k : integer;
r,t,p : double;
W := Form1.Image1.Width-1;
H := Form1.Image1.Height-1;
for k := 0 to H do
r := 1-k/H;
t := sqrt(r*(1-r)) - ArcSin(2*r-1)/2 + pi/4;
p := 2*t/pi*W;
with Form1.Image1.Canvas do
if k = 0 then MoveTo(Round(p),k)
else LineTo(Round(p),k);


Giving x(t) as a function of (t), visualized in a graph:

It is seen that the velocity becomes infinite near arrival.

Hope I've made no further mistakes this time.

Han de Bruijn

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