brentw (Brent Watts of Hickory, NC.
20001121)
How do I solve the following differential equations?
1) For the first DE, what you have is clearly the differential of [x sin(xy) + y]
so, the solution is: [x sin(xy) + y] = constant.
2) The second DE has a singular solution (the straight line x=0).
The general solution is obtained by noticing that the differential of
[x^{4}y^{3}
 3x^{3}y^{2}
+ x^{4}y^{2}]
is x^{2} times the given differential expression.
The solutions to the DE are therefore either curves of equation
[x^{4}y^{3}
 3x^{3}y^{2}
+ x^{4}y^{2}]such algebraic pieces and segments of the singular solution x=0
(joined at the points where they are tangent).
3) The last DE is a nobrainer, but a fairly annoying one.
The general idea is to make a linear change of variables so that the variables "separate"
easily.
For example, you may use x=u+v and y=(uv)Ö3,
which means
4u=x+yÖ3 and
4y=xyÖ3.
Your DE then becomes something like (please check):
[2(Ö3+3)v + Ö35] du = [2(Ö33)u + Ö3+5] dv
If I did manipulate things correctly (please double check),
this means that the solutions are the curves for which the following expression is constant:
Matthew A. (Yahoo!
20020721)
Singular Change of Variable
Solving the differential equation:
y dx = ( y e^{y}  2x ) dy
This example illustrates some of the issues which arise when
a given differential equation reduces to a tame equation
(here a simple quadrature) after a change of variable
which presents some singularities...
If y is not zero (a big "if")
we may replace x by the variable u = x y^{2 }.
du = y (y dx + 2x dy)
Multiplying both sides of the original equation by y turns it into:
Any value of the constant of integration C can be
used to obtain a solution over any domain in which y
does not vanish.
However, for a wellbehaved solution over a domain which crosses the line
y = 0, we must use
C = 2.
x =
[ (y^{2} 2y + 2) exp(y)  2 ] / y^{2}
It's only for this particular value of C that the
function x(y) is smooth about y = 0.
For small values of y, x(y) is indeed infinitely smooth:
Now, what about y as a function of x,
over the largest possible domain?
Well, the above function x(y) achieves a single negative minimum:
x = A = 0.169998...
when y = 1.45123...
For x > A, the solution has two branches.
One of those (y < 1.45123...)
has an asymptote at x = 0 and doesn't extend
to positive values of x.
The other branch (y > 1.45123...)
extends from x = A to
+¥
and it goes through x = 0,
y = 0 without the slightest glitch.
(20070730) Falling vertically in a fluid
(e.g., against air resistance).
The derivative of the downward speed is: v' =
g  ( 2 u v + v^{2 }) / l
In a vacuum, the downward speed
v increases at a rate g
equal to the acceleration of gravity.
The resistive forces exerted by the medium
(e.g., air) include a viscous resistance
(Stokes drag, proportional to v ) and a pressure term
dubbed quadratic drag,
proportional to the square of v.
With the notations introduced above, 2u is the speed at
which the quadratic drag becomes as large as the viscous term.
Loosely speaking, 2u marks the middle of a transitional
regime where the flow is no longer laminar, although the quadratic
term of the turbulent regime is not yet dominant
(2u is often dubbed "critical speed", although there's nothing
critical about it). The symbol
l is a length which is inversely
proportional to the magnitude of the pressure drag at high speeds.
Let's recast the above expression:
l dv/dt =
( lg + u^{2 }) 
( u + v )^{ 2}
Introducing a speed w > u defined by the relation
w^{2} = lg + u^{2 }, we obtain:
l^{ }dv
= ^{ }
[ w^{2}  (u+v)^{2} ] dt
Therefore :
2_{ }w dt
= _{ }
2 w l dv
_{ } = _{ }
l dv
_{ } + _{ }
l dv
w^{2 } 
( u + v )^{ 2}
w + u + v
w  u  v
This is readily integrated from the beginning of the fall
(v = 0) at time t=0 :
2 w t / l =
Log ( [w+u+v] / [wuv] ) 
Log ( [w+u] / [wu] )
Solving for v
(HINT: take the exponentials of both sides) we obtain:
v = ( w  u )
[_{ } 1 
2 w
_{ } ]
(wu) + (w+u) exp ( 2wt / l )
By integration, this yields the
height traveled since the beginning of the fall:
h = ( w  u ) t +
l Log
[_{ }
w + u
+
w  u
exp (  2wt / l )_{ }]
2 w
2 w
In practice, the exponential vanishes quickly and the
dropped object travels at a uniform
terminal velocity (wu) according to the following equation:
h = (wu) t  D
The distance D = l [ Log 2  Log (1+u/w) ]
can be interpreted either as a
time delay or as a correction in height
if you time the fall of an object dropped from an altitude Z.
The actual duration of the fall will be the time it would have taken
to travel the distance Z+D at a uniform speed equal to the terminal velocity.
According to the above, the terminal velocity (wu)
is the positive solution of the following quadratic equation:
v^{ 2} + 2 u v
 l g
= 0
That solution is best expressed by the following
robust formula (which never entails
any loss of precision in floatingpoint arithmetic, regardless
of the relative magnitudes of the parameters involved).
Terminal Velocity v = wu
v =
v_{0}
where (v_{0})^{ 2}
= l g
exp (argsh u/v_{0 })
This formula unifies and interpolates the following extreme regimes:
u << v_{0 } :
Pure pressure drag. Terminal velocity is
v = v_{0}
u >> v_{0 } :
Pure viscous resistance.
v = (v_{0 })^{ 3}/ u^{ 2}
Actual Physics :
The thrill of a closed solution to the above differential equation
does not make that differential equation physically meaningful
to begin with.
In fact, the two resistive terms in the equation correspond to
two separate speed regimes.
It turns out that each term is small in the regime where the
other is valid. Therefore,
adding up the two terms for transitional speeds is
mathematically appealing.
Such a nice approximation may lack a firm physical basis,
but I happen to like it.
A resistive force proportional to speed occurs in the lowspeed (viscous) regime
where there's absolutely no turbulence in the wake of the moving object
(a socalled "laminar flow"). In that case, a formula was established
by Sir George Stokes (18191903)
giving the resistive force F exerted on a sphere
of radius r moving at speed v in a fluid of
dynamic viscosity
h (Stokes' Law,
1851):
F = 6p h r v
On the other hand,
quadratic drag (proportional to v^{ 2 })
occurs in the turbulent regime normally associated with motion through air :
F = ½ r
(C S) v^{ 2}
= ½ r
(C p r^{ 2 }) v^{ 2}
In this,
S = p r^{ 2} is the crosssection
of the object, C is a dimensionless
drag coefficient (about 0.47 for a smooth sphere)
whereas r is the fluid's density.
Taking the above at face value for a sphere of mass m and
radius r, we may reconcile both equations with our original notations:
l =
2m / rCpr^{2}
and
u =
6h / rCr
For dry air at 20°C (1 atm)
r = 1.204 kg/m^{3}
and
h = 1.808 10^{5} Pa.s.
So, for a sphere (C=0.47) of radius r = 1 mm,
we obtain u = 0.1917 m/s.
A sphere of density d has a mass
m = d 4pr^{3}/3. So
l is proportional to r:
l =
r (8d / 3rC)
With the above numbers and
d = 1000 kg/m^{3}, we obtain
l = 4.7 r.
For steel ball bearings (d = 7850 kg/m^{3 })
we would have l = 36.9 r.
(20101212) Jet Propulsion in Outer Space
Rockets are propelled by expelling gases backward at speed u.
Let m and v be the mass and speed of a rocket (assumed to
move along a straight line).
As an infinitesimal mass dm of gases
is expelled at relative speed u
(or absolute speed vu), the conservation of momentum implies:
m dv + (vu) dm = 0
This means that m (vu) is a constant (= m_{o }u).
v = u (1  m_{o }/m )
(20120728) Riccatti Equation (Jacopo Riccati, 1720)
y' = y^{ 2} + a (x) y + b (x)
This is the standard form of a differential equation where the derivative of the
unknown (y) is a quadratic function of that unknown.
Any such equation can indeed be reduced to the above form, at least piecewise,
by changing the variable (x) so that the coefficients of y' and y^{ 2 }
are equal and can thus be factored out (on any interval of x where they don't vanish).
Introducing a new function z defined via y = z'/z we have:
z''/z + (z')^{2}/z^{2}
=
(z'/z)^{2} + a (x) (z'/z)
+ b (x)
Cancelling like terms and multiplying by (z) this boils down to:
0 = z''
 a (x) z'
+ b (x) z
This is just a secondorder linear (and homogeneous) differential equation.
Conversely, any homogeneous secondorder linear differential equation can be transformed into
a Riccati equation by the reverse process. Those two types of differential equations
are thus, essentially, of the same difficulty.
Note that the middle term can be eliminated by looking for z in the form
of a product of two functions z = k(x) u(x) which yields:
0 =
( k'' u + 2 k' u' + k u'' )

a (x) ( k' u + k u' )
+ b (x) k u
Let's choose for k a solution of the equation 2 k' = a (x) k .
We obtain:
0 = k u'' +
[ k'' 
a (x) k' + b (x) k ] u
Using k' = ½ a k
and
k'' = ½ a' k +
¼ a^{ 2} k this becomes: