YourDE is a separable differential equation. After taking the square root

it turns to du/(u-c}^(1/2)=dt or -du/{u-c}^(1/2)=dt and after integration

t=C1 +(u-c)^(1/2), or t=C1 -(u-c)^(1/2),

By differentiating both sides of the equation u'(t)^2=4(u(t)-c), we obtain u''(t)=2. So u(t)=t^2+at+b where a, b constants such that 4b-a^2=4c

almost - if u(t) is a constant with value c, it also satisfied the equation (after differentiating in your excellent solution, Latreuch assumed u'(t) was nonzero to cancel it out and get one family of solutions)

Your differential equation is implicit (not solved for u'). In such equations geometric singularities occur at which they may behave very differently from explicit equations. In your case, the situation is not so difficult. The whole solution space splits into a general solution and a singular solution which is an envelope to the general solution. The general solution consists of two families of parabola u = t^2+/-sqrt(b-c)*t+c where the parameter b must satisfy b>=c. The singular solution is the straight line u = c and is the tangent to the parabola at the point t=+/-sqrt(b-c).

My answer is solved by the mathematical site http://www.wolframalpha.com/ and it is:

The given differential equation is NOT a Ricatti equation. A Ricatti equation is of the form u'(t) = a(t)+b(t)u(t)+c(t)u^2(t), i.e. it is a standard ODE with a quadratic right hand side. The problem with Alireza's equation is the fact that it is not solved for the derivative u' which allows for additional phenomena like singular solutions typically not treated in standard textbooks on ODEs. Because of the simple form of the equation it is indeed not difficult to come up with the general solution (as shown above, even software like Wolfram Alpha is able to do it), but most of the answers so far overlooked the additional singular solution!

I'm not agree that 2 equations: (x(t)^2)(1-x'(t))^2=c and u'(t)^2=4(u(t)-c) are equivalent after the substitution: u(t)=x(t)^2 and u'=2xx' . My answer is solved by the mathematical site http://www.wolframalpha.com/ and it is below. About the function Product Log please, see the same site.

Dear Peter T. Breuer,

It is evidently that (x(t)^2)(1-x'(t))^2=c after the substitution: u(t)=x(t)^2 and u'=2xx' will has the sort u'(t)^2/4-(+)u' (t)sqrt(u)+u=c, which is different of u'(t)^2=4(u(t)-c) . But ODE (x(t)^2)(1-x'(t))^2=c has a simply solution, because: mod(x(t)*(1-x'(t))=sqrt(c) is resolvable about t. I think that we have to prefer the solution of http://www.wolframalpha.com/ . Please, answer me. Your sincerely, Anna Tomova

Please, may I know what is happening? Why are there errors in almost all the answers given so far to the question by Alireza Ahmadi? If it is understood that x and u are functions of t then x(t) and u(t) should be written simply as x and u respectively.

The 1st order nonlinear ODE is x^2(1 – x')^2 = c. I hope I got this right.

Applying the transformation u = x^2, u' = 2xx' the ODE becomes

(x - xx')^2 = c,

[sqrt(u) – u'/2]^2 = c

Solving for u' I obtain u '= 2[sqrt(u) +/- sqrt(c)] or (u')^2 = 4[sqrt(u) +/- sqrt(c)]^2

Where +/- means plus or minus.

Consequently, the transformed ODE (u')^2 = 4[u – c] given by Ahmadi is wrong. Anna Tomova had earlier pointed this out.

The original ODE doesn’t require any transformation to obtain its solution. It is variable seperable and can be solved directly as follows:

x^2(1 – x')^2 = c, 1 – x' = +/- sqrt(c)/x

x' = dx/dt = 1 +/- sqrt(c)/x

This can be expressed in variable-seperable form as

dt = xdx/[(x +/- sqrt(c)]

Integrating this ODE yields

t = A + x – sqrt(c) ln [x +sqrt(c)]

or t = A + x + sqrt(c) ln [x -sqrt(c)]

which are the solutions of the ODE and A is a constant to be determined and ln is natural logarithm. As can be seen x cannot be expressed explicitly in terms of t.The solution is asymptotic to the straight line t=x+A at infinity.

Dear Peter T. Breuer and Amaechi J. Anyaegbunam ,

I send my answer to You as a file attachment. Please, answer me, Your sincerely, Anna Tomova

Dear Anna V. Tomova:

I have studied your manuscript and have not really discovered an alternate solution to the ODE. You have done lengthy calculations and arrived at the same solutions I gave in my previous post. Unfortunately, you have refused to agree that my own solution is correct .

You have just introduced Wolfram's function W(f) as the solution to the transcendental equation:

ze^z = f(t) (1)

i.e. z = W(f) is a solution to the above.

Now, W(f) is neither an elementary function nor a widely recognized higher function like J_v(x), Gamma function, Beta function, Legendre polynomial etc. Wolfram may have just introduced it in its repertoire of functions. But we have to ask ourselves if Eq. (1) occurs so frequently in applications to warrant its having a function invented for it. Proof of my assertion is that the function W(f) is not widely known in the Mathematics community nor do I know of any tabulations for it.

Hence, I am in order for saying the solution of the ODE has no format in which x can be expressed explicitly in terms of t.

Thank you!

Dear Peter T. Breuer ,

Why You don't read my attachment.doc? Please, write my in what format You can read files attachment and I will converter my attachment. Please, answer me, Your sincerely, Anna Tomova

Dear Amaechi J. Anyaegbunam ,

I' m very happy that You answered my so quickly. Of course, Your solution is right and I apologize, that I missed to congratulate You about it. It is evidently that my transformations relate to the same solutions. But I'm not agree that the function W(z) hasn't an interest for the mathematicians. I think that it is introduced for the solutions of many indefinite integrals too. Thank You very much. Your sincerely, Anna Tomova

Peter Breuer is quite right. The ODE has a definitely obvious singular solution x = sqrt(c).

The function W(z) is quite obscure. My electronic calculator doesn't have it. Fortran, Matlab, C and C++ do not have it. We have to wait for some time in future when W(z) will become popular.

Dear Peter T. Breuer and Amaechi J. Anyaegbunam, ·

We can find the whole information for the function W(z) at the address: http://en.wikipedia.org/wiki/Lambert_W_function etc. My opinion is that the yang author with the address: https://www.researchgate.net/profile/Alireza_Ahmadi was able to find one ODE: (x(t)^2)(1-x'(t))^2=c, in deciding which participates the function of Lambert. Lambert first considered the related Lambert's Transcendental Equation in 1758,[3] which led to a paper by Leonhard Euler in 1783[4] that discussed the special case of wew. However the inverse of wew was first described by Pólya and Szegő in 1925.[5][citation needed] The Lambert W function was "re-discovered" every decade or so in specialized applications but its full importance was not realized until the 1990s. When it was reported that the Lambert W function provides an exact solution to the quantum-mechanical double-well Dirac delta function model for equal charges—a fundamental problem in physics—Corless and developers of the Maple Computer algebra system made a library search to find that this function was in fact ubiquitous to natur. Your sincerely, Anna Tomova.

Dear Peter Breuer the point Anna Tomova is making is that by using the Lambert function W(z) the solution of the ODE can be expressed in the form x = x(t)

Consider one of the solutions

x + sqrt(c) ln [x - sqrt(c)] = t + a

This can be rewritten as

x/sqrt(c) -1 + ln [x - sqrt(c)] = (t+a)/sqrt(c) - 1

Taking exponentials of both sides and dividing by sqrt(c) it is obtained that

[ x/sqrt(c) -1] exp [x/sqrt(c) - 1] = [1/sqrt(c)] exp[ (t+a)/sqrt(c) – 1]

Noting that z = x/sqrt(c) -1

the solution can then be written as

x/sqrt(c) -1 = W{[1/sqrt(c)] exp[ (t+a)/sqrt(c) – 1] }

or x = sqrt(c) + sqrt(c) W{[1/sqrt(c)] exp[ (t+a)/sqrt(c) – 1] }

But my contention is that the function W(z) is not important because I can easily set up a Newton Raphson routine to determine the roots of W exp(W) = z.

W(z) is not really worthy to be established as a special function in its own right except for it symbolic use to allow W, the solution of W exp(W) = z to be written as W(z).

Thank you

Bessel's ODE arises naturally from many physical problems such as the solution of axisymmetric vibration of a circular membrane; asixsymmetric steady state flow of heat through a circular disc etc

Legendre's ODE and polynomials arise in the solution of the steady state spherical-symmetric flow of heat through a sphere and other BVPs involving the sphere.

Elliptic integrals occur in the solution of steady state ground water seepage under dams and through finite soil layer and in electric potential distribution problems.

In what physical applications do the ODE (x+a)x' = x occur on a regular basis in science and engineering? A one off application of W(z) for the "exact solution to the quantum-mechanical double-well Dirac delta function model for equal charges" does not warrant its further consideration.

Dear Alireza Ahmadi

We have not heard from you on Research Gate concerning the question you have asked.

I wish to seal my contribution.

It has been shown that the solution of the ODE x^2(1 – x')^2 = c is

t +a = x + sqrt(c) ln [x - sqrt(c)]

where a is an arbitrary constant. The other solution is obtained by adding a minus sign to sqrt(c)

The solution can be expressed, after Anna Tomova, alternatively as

x = sqrt(c)[ 1 + W(f)]

where f = [1/sqrt(c)] exp[ (t+a)/sqrt(c) – 1]

and W(f) is the solution of Wexp(W) = f

The following accurate approximations could be written for W(f):

0

I send my answer to You as a file attachment with any corrections.. Please, answer me, Your sincerely, Anna Tomova

I send my answer to You as a file attachment with the additions for all possible 4 cases. Please, answer me, Your sincerely, Anna Tomova

Dear Anna V. Tomova.

I don't understand the type of answer you expect from me or from Peter T. Breuer

It is agreed that the given ODE (x(t)^2)(1-x'(t))^2 = c has the two solutions below wherein t is expressed explicitly in terms of x.

t = A+ x + sqrt(c) Ln [x - sqrt(c)] ----------------------- (1)

t = B + x - sqrt(c) Ln [x + sqrt(c)] ----------------------- (2)

where Eq. (1) is valid in the real domain provided x > sqrt(c) and Eq. (2) is valid provided x > -sqrt(c)

However, you seem to be arguing that x can be expressed explicitly in terms of t if, the rarely used and novel Wolfram function, W(f), is introduced.

But, my contention, supported by Peter Breuer, is that the use of W(f) is not a necessity because

(a) x can be calculated for given t using Eq. (1) or (2) with a suitable computer program that uses Newton Raphson iteration.

(b) W(f) is not a tabulated function that is widely available.

W(f) is implemented by Mathematika software, hence it requires a computer for its utilization. Thus, it has no advantage over the alternative method of solving for x using Newton-Raphson iteration. Besides not every researcher owns Mathematika.

Intuitively Eq. (2) seems more appropriate. Assuming naturally that x = 0 when t= 0 it is obtained that B = sqrt(c)Ln(sqrt(c)) so that

t = x – sqrt(c)Ln[x/sqrt(c)+1] ------------------------------------- (3)

For a more stable iteration this is best expressed as

f(x) = exp[(x-t)/sqrt(c)] – [1+x/sqrt(c)] = 0 ---------------------- (4a)

f '(x) = f/sqrt(c) + x/c ------------------------------------------ (4b)

The very simple and short Fortran program for determing x given t is reproduced below. It obtains x for t = 0.001, 0.01, 0.1, 1.0 10.0, 100.0, 1000.0, 10000.0. The answers are x = 5.8863, 18.6188, 58.9233, 186.7882, 595.2517, 1928.4648, 6570.3710, 25796.9327. The value of c was taken to be the speed of light 3*10^8 m/s.

PROGRAM ITERATION

IMPLICIT NONE

REAL*8 f, derf, x, x2, c, eps

REAL t, fr, derfr, xr, tv(100)

! tv = vector of times

INTEGER i, j, num, yes

PARAMETER (c = 3.0d8, eps = 1.0d-14)

tv(1:8) = (/.001,.01, .1, 1., 10., 100., 1000., 10000./)

num = 8

DO j = 1, num ! 1st DO

WRITE(*,*) ; WRITE(*,*)

t = tv(j)

! Commence iteration for value of x corresponding to t

x = sqrt(2.*t*sqrt(c)) ; i = 1

14 f = exp((x-t)/sqrt(c))-(x/sqrt(c)+1)

derf = f/sqrt(c) + x/c

x2 = x - f/derf

xr = REAL(x2); fr = REAL(f); derfr = REAL(derf)

WRITE(*,*) ' x = ',xr,', f = ',fr,", f' = ", derfr

IF (abs(f)

Is there a reason you don't want to use:

u(t) = t^2 + uo + 2 t sqrt(uo-c) ?

or

u(t) = t^2 + uo - 2 t sqrt(uo-c)

where uo = u(0) ?

Sorry if I misread the problem.

Alireza Ahmadi wanted to solve the 1st order nonlinear

ODE x^2(1 – x')^2 = c --------------------- (1)

and proposed the transformation

u= x^2, u' = 2xx'. When this transformation is applied the

correct result is u '= 2[sqrt(u) +/- sqrt(c)] -------------- (2)

The transformed ODE (u')^2 = 4[u – c] ----------------- (3)

given by Ahmadi is wrong. This has been pointed out

in previous posts

Hence, it is incorrect to solve Eq. (3) as a prelude to

solving Eq. (1). As was also pointed out previously

no transformation is needed before solving Eq. (1)