easier solution for this ODE?
Grigorios Kostakos
y''(x) - 7y'(x) + 12y(x) = 0,
after the substitution z=dy/dx, we have
z(dz/dy)-7z+12y = 0.
For z=/=0 we have
(dz/dy) - 7 + 12/(z/y) = 0
and after the substitution w=z/y we have
y(dw/dy) = 7- 12/w -w
which by separation of variables gives
(w/(7w-12-w^2))(dw/dy) = 1/y
and by intergration leads to the expression
(w-3)^3 = cy((w-4)^4), where c=/=0 is constant.
So the problem is that for the solution I have to solve a non-easy
quadratic equation...
The question: is there any easier way to solve this ODE?
Thanks
Grigorios Kostakos
Bill Branson
assume that y = e^(rx), substitute this into your original
equation, and see what conditions r must satisfy...
--bill
Dr. Richard L. Hall
"Grigorios Kostakos" <grig...@pathfinder.gr> wrote in message
news:35eea5fc.03070...@posting.google.com...
> for the solution of ODE
> y''(x) - 7y'(x) + 12y(x) = 0,
This is a linear DE with constant coefficients. The solutions should be of
the form y = exp(ax). Substitute this into the equation and solve for a.
That will give you two roots a1 and a2. So the general solution will be
y = C1 exp(a1 * x) + C2 exp (a2 * x)
where C1 and C2 are constants usually determined by the initial conditions.
David Ziskind
<35eea5fc.03070...@posting.google.com> addressed the subject
of integrating the 2nd order linear DE with constant coefficients, by a
quadrature approach.
While the trial solution method is almost universally employed for this
ODE, in this particular case, quadrature is normally possible and has
some advantages.
We assume:
1. G'' + b*G' + c*G = 0 on an open interval (a,b). At this point,
we make no assumption as to whether or not there exists a G such that
this relation holds true. The forthcoming deduction will tell us what
happens if there is.
2. G is continuous
3. G' exists and is continuous
5. G'' exists
6. Let d =def sqrt(b^2 - 4*c)
7. Let r =def -b/2 + d/2
8. There exists at most a finite number of points x in (a,b) such
that G'(x) = r*G(x) . From here on, we will refer to such points as
“special points”.
9. Let F be defined implicitly by: G(x) = F(x)*exp(r*x)
10. d ~= 0
11. Let x1, x2, x3, ... be the sequence of points on the real axis
obtained as follows: 1st, x1=a. 2nd, move to the right and let x2 be
the first special point encountered. Continue in this fashion -- the
next special point is labeled x3, etc. Eventually, b will be
encountered and will become xn for some n. Note that a (and/or b)
might be special points.
From these premises 1 through 11, we have the following deduction. In
the following, we introduce two new functions, which are G and F with
their domains restricted to the i th subinterval, that is, the open
interval (xi, x(i+1)). Rather than introduce new nomenclature, we will
(temporarily) refer to the restricted G simply as G, and similarly for
F.
The function x->exp(r*x) will be denoted E .
12. r^2 + b*r + c=0 by direct computation using (6) and (7)
13. Now substitute into (1) using (9), and obtain:
F''*E + F'*(2*E'+b*E) + F*(E''+b*E'+c*E) = 0
14. By 12, the term multiplying F is 0. We may divide by E since this
is never zero. The coefficient of F' is 2*r + b which by (7) is d. We
have:
15. F'' + F'*d = 0
16. By use of (9) again, we find that F' = (G'-r*G)/E . By
construction of the current subinterval, each interior point is not a
special point. Therefore, F' ~= 0 within the subinterval. In (15),
transpose and divide by F'. Obtain:
17: (F')'/F' = -d This is easily integrated, and
18. ln(F'(x)) = -d*x + c1 c1, c2, ... will be constants of
integration.
19. F'(x) = c2*exp(-d*x)
20. Integrate a second time. The integral of exp(-d*x) has two
different forms, depending on whether or not d is 0. In this case, by
assumption 10, d is not 0, and the integral of exp(-d*x) is
-(1/d)*exp(-d*x) + constant. Apply (9), and obtain:
G(x) = [c3*exp(-d*x) + c4] * exp( ((-b/2)+(d/2))*x )
21. When we multiply the outside exp term into the sum, we obtain the
classic sum-of-two-exponentials form, with each exponential coefficient
a characteristic root:
G(x) = c3*exp( ((-b/2)-(d/2))*x )
+ c4*exp( ((-b/2)+(d/2))*x )
This completes the calculation of G on the i th subinterval. By
performing the same calculation for all subintervals, and using the
continuity assumptions for G and G', we can take limits, solve a system
of two equations which has a non-zero determinant, and show that c3,c4
to the left must equal c3,c4 to the right. The constants of
integration are equal for all subintervals. Equation 21 holds
throughout the entire interval (a,b), including the special points.
Comments:
a. If d=0, the integration at (20) gives F(x) = c2*x + c5, and thus,
G(x) = (c2*x + c5)*exp(-b*x/2).
b. The assumption of at most, a finite number of special points is
specific to the quadrature approach. (I do not see a convincing way it
can be avoided.) In any event, however, it is not required for the
trial solution method (but see d.3 below).
c. One can take the function appearing on the right side of (21) and
show by direct substitution that it satisfies (1) -- that is, Equation
1 does indeed have a solution.
d. The point of the quadrature exercise can be stated as follows:
--First, there is the question of how one finds the trial solutions in
the first place. In this case, quadrature reveals them.
--Second, there is always the possibility that a fortuitously found
trial solution (or set) is in fact, NOT fully general. This is
actually illustrated here by the case d=0. Another example is the case
of ODE's with envelopes. Quadrature when it is available tells us
absolutely that the functions found are the ONLY ones that can be
solutions.
--Third, the trial solution method only makes complete sense when we
have some kind of exterior uniqueness result that tells us that the
given ODE with initial/boundary conditions has at most one solution.
If so, a solution which is some combination of trial solutions is in
fact valid. However, it takes some work to develop a uniqueness
theorem. Quadrature provides a method of sidestepping this difficulty.
David Ziskind
zis...@ntplx.net