| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1186.1 | I think this looks right... | AKQJ10::YARBROUGH | I prefer Pi | Wed Jan 31 1990 16:25 | 19 |
| > Solve
> y'' = -k/(y**2)
MAPLE:
> solve(diff(diff(y(x),x),x)=k/y^2,y);
1/2
y
int(---------------------, y) = x + C1,
1/2 1/2
2 (- k + C y)
1/2
y
int(- ---------------------, y) = x + C1
1/2 1/2
2 (- k + C y)
|
| 1186.2 | Mechanical advantage!! | ATPS::DAWSON | | Wed Jan 31 1990 16:36 | 1 |
| Does MAPLE get the cookie?
|
| 1186.3 | I'll give it a maple cookie if it wants one... | DWOVAX::YOUNG | If it bleeds, we can kill it. | Wed Jan 31 1990 20:21 | 7 |
| Looks good, help me with the MAPLE formatting here though.
Are these 2 distinct solutions? And what are C and C1 ? And does the
fact that it doesn't go any farther mean that these are not solvable
Integrals?
-- Barry
|
| 1186.4 | | ALLVAX::ROTH | It's a bush recording... | Thu Feb 01 1990 07:25 | 6 |
| I don't have MAPLE online right now, but shouldn't you have used
dsolve(explicit) for the solution?
The solution doesn't look right.
- Jim
|
| 1186.5 | It's not all that great... | AKQJ10::YARBROUGH | I prefer Pi | Thu Feb 01 1990 14:59 | 21 |
| The cookie must, of course, be made with Maple sugar...
> Are these 2 distinct solutions?
Yes. They only differ by sign.
>And what are C and C1 ?
Arbitrary constants of integration.
>And does the fact that it doesn't go any farther mean that these are not
>solvable Integrals?
Not necessarily - only that MAPLE couldn't further reduce them without
help.
RE .4; Yes, if you have MAPLE V4.1+ available, dsolve is the tool to use.
The results I gave were obtained with MAPLE 3.3. I have 4.1 installed on
another system but it doesn't work on D.E.'s because a library function,
exp.m, is missing. !@%^&*(. Have to bug WATCOM about that. Version 4.3,
where are you?!?!?
Lynn Yarbrough
|
| 1186.6 | A particular solution to work with... | CSC32::JAGGER | | Thu Feb 01 1990 19:41 | 14 |
| I can give you one particular solution if you want it.
if Y = f(x) and lim x -> 0, y -> 0 for x > 0 then
if Y=a*x^n, I can solve for n and a and I get
Y= (9/2*k*x^2)^(1/3) where ^ is exponentiation.
this solves Y'' = - k/ Y^2, unfortuneatly this equation is
non-linear so I am not too sure how usefull this solution is.
Also, there may be many solutions with x=0, Y=0.
TOM
|
| 1186.7 | Last reply found by letting C=0 in .1 | CSC32::JAGGER | | Thu Feb 01 1990 19:54 | 13 |
| I should have looked at MAPLE'S solution carefully (see .1) It does provide
a particular answer for any c and c1, if I set c=0 I get my answer,
that I provided in my last reply. (I have never used MAPLE)
Also, it shows that the solution must be invariant with respect to
the independant variable, since the variable does not appear in the
equation. As maple shows:
if Y=f(x) is a solution so is Y=f(X+C1). So I can generalize
my particular solution...
TOM
|
| 1186.8 | solution of y''=-k/(y**2) | DCC::PARETI | | Mon May 28 1990 06:15 | 43 |
|
Seeking the solution of
y'' =-k/(y**2)
put dy/dx = z (1)
then follows dz/dx = -k/(y**2) (2)
Divide (1) and (2) :
dy/dz = -z*(y**2)/k or
dy/(y**2)= -z*dz/k (3)
Integrating (3) yields :
-1/y = -(z**2)/(2*k) (4)
replace z in (4) with dy/dx :
dy/dx=(2k/y)**1/2
which can be rewritten as :
((2*k)**(-1/2) ) * dy * y**0.5 = dx (5)
Integrating (5) gives :
( (2*k)**(-1/2) ) *(2/3) * y**(3/2) = x + constant
the latter can be rewritten in the form y=y(x) :
y = [ (3/2)* (2*k)**(1/2) ]**(2/3) * (x + constant)**(2/3) (6)
y' = [ (3/2)* (2*k)**(1/2) ]**(2/3) * (2/3)*(x + constant)**(-1/3) (7)
y''= -(1/3)*(2/3)*[ (3/2)* (2*k)**(1/2) ]**(2/3) * (8)
(x + constant)**(-4/3)
y**2 = [ (3/2)* (2*k)**(1/2) ]**(4/3) *(x + constant)**(4/3) (9)
It is easy to prove from (8) and (9) that y''*y**2 = -k
|
| 1186.9 | | GUESS::DERAMO | that Colorado Rocky Mountain high | Tue May 29 1990 16:26 | 11 |
| re .8
>> dy/(y**2)= -z*dz/k (3)
>>
>> Integrating (3) yields :
>>
>> -1/y = -(z**2)/(2*k) (4)
Plus an arbitrary constant.
Dan
|
| 1186.10 | applying .9 to .8 -- a more general form | TRACE::GILBERT | Ownership Obligates | Wed May 30 1990 13:57 | 38 |
| > dy/(y**2)= -z*dz/k (3)
>
> Integrating (3) yields :
>
-1/y = -(z**2)/(2*k) + C (4)
replace z in (4) with dy/dx :
dy/dx = sqrt(2k(C+1/y)) (may be plus or minus)
which can be rewritten as :
(2k(C+1/y))**(-1/2) * dy = dx (5)
Integrating (5) gives (below, w = k(C+1/y), and D is a constant):
if Ck > 0 :
sqrt(w) log(sqrt(w)-sqrt(Ck)) - log(sqrt(w)+sqrt(Ck))
(6a) ----------------- + ---------------------------------------------
sqrt(2) C(w - Ck) 2 sqrt(2) C sqrt(Ck)
= x + D
if Ck = 0 :
sqrt(w) 1
(6b) ----------------- - ----------------- = x + D
sqrt(2) C(w - Ck) sqrt(2) C sqrt(w)
if Ck < 0 :
sqrt(w) atan( sqrt(w)/sqrt(-Ck) )
(6c) ----------------- + ------------------------- = x + D
sqrt(2) C(w - Ck) sqrt(2) C sqrt(-Ck)
It's not easy to rewrite any of these in the form y=y(x).
|
| 1186.11 | | TRACE::GILBERT | Ownership Obligates | Fri Jun 01 1990 01:25 | 8 |
| re .10:
> if Ck = 0
The equation given for this case is totally bogus.
That's what comes from unequivocally accepting a symbolic
math package's results. The right answer in this case
is given by Pareti in .8.
|
| 1186.12 | MAPLE V attack | CIVAGE::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Mon Oct 21 1991 13:40 | 22 |
| MAPLE V does a better job at solving this (especially given a more
disciplined description of the problem!). In the following,
diff(...,x$2) means the second derivative w/r/t x:
> dsolve(diff(y(x),x$2)=k/(y^2),y(x));
1/2 1/2 1/2
%1 k ln(2 _C1 %1 + 2 _C1 y(x) - 2 k)
x = - ----- - --------------------------------------- - _C2,
_C1 3/2
_C1
1/2 1/2 1/2
%1 k ln(2 _C1 %1 + 2 _C1 y(x) - 2 k)
x = ----- + --------------------------------------- - _C2
_C1 3/2
_C1
2
%1 := - 2 y(x) k + _C1 y(x)
Solving this mess for y(x) is still a little daunting...
|