Finding the equation from the solution
sert
is:
(a^2)*y +2*b*x -5 = 0
with y=f(x) and a,b constants?
Dave L. Renfro
This is an equation of a line, so y'' = 0.
Alternatively, differentiate until the constants
are removed:
(a^2)*y' + 2b = 0
(a^2)*y'' = 0
y'' = 0
Sometimes this second method gives you an equation
of higher order than you should have. However, since
there are two constants in your original equation,
an ODE of order 2 (what we got) is what we want.
This method is called "elimination of constants" in
older calculus and ODE texts. More recent ODE texts
seem to have greatly reduced the amount of attention
devoted to this topic.
http://books.google.com/books?q=differential-equation+elimination-of-constants
Dave L. Renfro
sert
news:16129610.1209150721...@nitrogen.mathforu
m.org:
> (a^2)*y' + 2b = 0
>
> (a^2)*y'' = 0
>
> y'' = 0
So the answer is y'' = 0 ?
Isn't that a bit too general? That has any line as a solution,
not just the ones that come from
Dave L. Renfro
> So the answer is y'' = 0 ?
>
> Isn't that a bit too general? That has any line
> as a solution, not just the ones that come from
>
> (a^2)*y +2*b*x -5 = 0
I don't think you're going to get a simple
set of ODE-like conditions that pin things down
to your form (coefficient of y is non-negative
and additive constant of -5; the '2' with the
'b' is no concern, as any number is twice
another number, and vice-versa), but maybe
someone else will have an idea.
Where did this problem come from? I just assumed
it was a problem from an ODE text.
Dave L. Renfro
quasi
wrote:
Why not just differentiate and solve for y'?
quasi
Jon Slaughter
"sert" <je...@hotmail.com> wrote in message
news:Xns9A8BDC9D533...@147.102.222.230...
there are an infinite number of differential equations with this solution.
You need to be more specific.
obviously if f(x,y) = a^2*y + 2bx - 5
then f_x = 2b and f_y = a^2
so f_x + f_y = 2b + a^2 is one equation(you'll need an IC/BC to get f(x,y) =
0)
or if you write
y(x) = 6/a^2 - 2b/a^2*x
then
y' = -2b/a^2
which is another DE
Not sure why you want to find "the differential equation". To find a
differential equation it only involves taking derivatives and combining them
in various ways to get some type of differential equation that works.
another example:
since y'*y = 4b^2/a^4*x - 12b/a^4
it is another differential equation.
cos(y') + y^2 = f(x) is another(obviously f(x) is the express on the LHS in
terms of x)
but whats the point?
remember, how do you validate the solution of a differential equation?
sert
org:
> Where did this problem come from? I just assumed
> it was a problem from an ODE text.
>
It's from an ODE textbook.
It has a part about finding the DE that has a parametric family
of curves as the solution. It says you should differentiate as
many times as you have constants. I did that and did find y''=0
of course but I thought I might be missing something because it
didn't make any sense to give something like
(a^2)*y +2*b*x -5 = 0
when any line would give the same result. Looks like I'm not
missing anything. Thanks for the help.
sert
news:NHqQj.487$To6...@newssvr21.news.prodigy.net:
> You need to be more specific.
>
The idea is to find a DE without the constants. Looks like the
only suitable DE then is y''=0.
Weird way to put a problem though, wouldn't you agree?
G.E. Ivey
Robert Israel
> Jerry Sargent wrote:
>
> > So the answer is y'' = 0 ?
> >
> > Isn't that a bit too general? That has any line
> > as a solution, not just the ones that come from
> >
> > (a^2)*y +2*b*x -5 = 0
>
> I don't think you're going to get a simple
> set of ODE-like conditions that pin things down
> to your form (coefficient of y is non-negative
> and additive constant of -5; the '2' with the
> 'b' is no concern, as any number is twice
> another number, and vice-versa), but maybe
> someone else will have an idea.
Since you can always multiply the equation by a
positive constant, there's nothing special about -5.
You can prevent the coefficient of y in
c y + 2 b x - 5 = 0 from being negative, though: e.g.
y''/(|y - x y'| + y - x y') = 0
(since if c < 0 the denominator will be 0 and the equation
is undefined)
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Dave L. Renfro
> The idea is to find a DE without the constants.
> Looks like the only suitable DE then is y''=0.
>
> Weird way to put a problem though, wouldn't you
> [= Jon Slaughter] agree?
I'm note sure about Jon Slaughter, but it seems
weird to me. One of the uses of this is to obtain
simple equations that can be used to describe certain
useful parametrized classes of functions. The following
text, which I've taught from three times, has many
interesting examples and exercises on this topic near
the beginning of the book:
Earl D. Rainville and Phillip E. Bedient, "Elementary
Differential Equations", 7'th edition, Macmillian
Publishing Company, 1989.
Recently I came across a neat nested differential
expression for all conic sections. I don't remember
what it was now or where I saw it (I think it may
have been an American Mathematical Monthly problem),
but if I think about it when I get home, I might see
if I can look it up (I know where I jotted it down)
and post it tomorrow.
Dave L. Renfro
quasi
<georg...@gallaudet.edu> wrote:
>> On Fri, 25 Apr 2008 18:41:11 +0000 (UTC), sert
>> <je...@hotmail.com>
>> wrote:
>>
>> >How can we find the differential equation to which
>> the solution
>> >is:
>> >
>> >(a^2)*y +2*b*x -5 = 0
>> >
>> >with y=f(x) and a,b constants?
>>
>> Why not just differentiate and solve for y'?
>
> Because differentiating ONCE will not get rid of BOTH constants.
Ok, I follow.
The desired differential equation is not allowed to be parametrized by
a,b.
quasi
Dave L. Renfro
> Recently I came across a neat nested differential
> expression for all conic sections. I don't remember
> what it was now or where I saw it (I think it may
> have been an American Mathematical Monthly problem),
> but if I think about it when I get home, I might see
> if I can look it up (I know where I jotted it down)
> and post it tomorrow.
The problem came from an Oxford University (exit?)
exam (for mathematics majors, I presume), unknown
date except it was before 1897 (probably at most
a few years before 1897). The reference I have for
the problem is in The Mathematical Gazette: stated
in Volume 1, Number 9 (October 1896), p. 67; solution
(no attribution is given) in Volume 1, Number 10
(February 1897), pp. 94-95.
Below is the problem and solution, edited a bit
from the original. After this are some examples I
found in handwritten class notes of mine from
Spring 1997 (used in classes I taught the Spring
semesters of 1997,1998, 1999, and 2001).
---------------------------------------------------
PROBLEM: Show that, by eliminating constants, the
general conic equation can be described by the
following differential equation:
(d^3 / dx^3) { [ (d^2 / dx^2)(y) ] ^ (-2/3) } = 0
or
[(y'')^(-2/3)]''' = 0
The left hand side is the third derivative of the
-2/3 power of the second derivative of y.
SOLUTION: The general conic satisfies
ax^2 + bxy + cy^2 + dx + ey + f = 0,
where a, b, c, d, e, and f are constants.
STEP 1: Solve for y in terms of x.
We have the quadratic equation
(c)y^2 + (bx + e)y + (ax^2 + dx + f) = 0,
whose solution (quadratic formula) is
y = (-bx - e)/(2c) +/- sqrt(Ax^2 + Bx + C)
It's clear from the quadratic formula that
what's under the square root will be a
quadratic polynomial in x. A lot of work
can be saved by not worrying about what
A, B, and C are in terms of the original
constants, since this won't be needed
later on.
STEP 2: Take the derivative of both sides
with respect to x.
y' = -b/(2c) +/- (Ax + B/2) / sqrt(Ax^2 + Bx + C)
STEP 3: Take another derivative of both sides.
The result, after some algebraic cleaning, is
y'' = +/- (AC - B^2/4)(Ax^2 + Bx + C)^(-3/2)
(The "algebraic cleaning" part is the only place
where the computations aren't essentially effortless.)
STEP 4: Isolate on one side of the equation a
quadratic expression involving x and then
take three successive differentiations with
respect to x.
(y'')^(-2/3) = (constant)(Ax^2 + Bx + C)
[(y'')^(-2/3)]''' = 0
---------------------------------------------------
Here are some examples I found in old notes of mine.
They are worked out in my notes, but I'll just
give the statements and answers here.
WRITE A DIFFERENTIAL EQUATION OF LEAST ORDER THAT
INCLUDES ALL CURVES FROM THE GIVEN FAMILY OF CURVES.
1. Lines through the origin.
xy' - y = 0
2. Lines whose slopes equal their y-intercepts.
(x+1)y' - y = 0
3. Circles centered at the origin.
yy' + x = 0
4. Circles with centers on the x-axis.
yy'' + (y')^2 + 1 = 0
5. Circles tangent to the x-axis.
[1 + (y')^2]^3 = [1 + (y')^2 + yy'']^2
6. Circles with centers on the line y = -x
and which contain the origin.
y' = [y^2 + 2xy - x^2]/[-y^2 + 2xy + x^2]
Dave L. Renfro