I have some information that can help you. First of all, try to see a good information web about differential equation (see the link).

And the another possibility is using Maple program. In my mind, this is better program for solving differential equations.

http://eqworld.ipmnet.ru/index.htm

I can get the analytical solution of your problem, but in the parametrical form x=f (p), y-g (p), where p=dy/dx.

Further it is possible to investigate this solution graphically having eliminated parameter p from the parametrical solution. I so arrive in my problems.

However I have to tell that the analytical solution will be very complex.

Vasiliy Belozyorov

I can get the analytical solution of your problem, but in the parametrical form x=f (p), y=g (p), where p=dy/dx.

Further it is possible to investigate this solution graphically having eliminated parameter p from the parametrical solution. I so arrive in my problems.

However I have to tell that the analytical solution will be very complex.

--

Vasiliy Belozyorov

Dear Colleague,

The solution - analytical - to your problem is to be found in the following classical book:

Erich Kamke, Differential equations. Solving methods and solutions. 

The book has known 6 editions, all in German. I have its Russian translation version. Hopefully there should be some English version also. I am able however to give you some hints:

1) You must write y explicitly as follows:y = Ex(dy/dx-(dy/dx)-1) + 2C , assuming that dy/dx is different of 0 on the interval of existence of the solution (otherwise your problem is trivial i.e. x=0)

In this way you will obtain a Lagrange d'Alembert type equation.

2) See paragraph 4.19 of the cited book of Kamke and follow the algorithm given there. You will be led to the analytic solution. However, as colleague Belozyorov suggests, it might be either quite complicated or given in parametric form.

GOOD LUCK!

Dear Ababu, it doesn't seem complex as it looks. I would model the question as a quadratic equation, meaning that I let (dy/dx)=z, and integrate it explicitly to get z=y/x. The original question will then look like: Exz^2+(2C-y)*z-Ex=0; the next thing would be to substitute for z with y/x to get y^2(E-C)+2Cy-Ex^2=0. Let E-C=k, so that the whole thing looks like: ky^2+2Cy-Ex^2=0; Solve again for y using the general quadratic formulae. you will get something like: y=(-2C+or- sqrt(4C^2+4kEx^2))/2k. You will realized that the only unknown is x which I assume if plotted as a variable like time or length or anything you will get the subsequent values of y since you know k, E and C. I also found out that your initial condition y(0) hold for this system but only if you take the positive value of the term with square root. If you take the negative value your initial condition will not hold and you will find that at x=0, y=-4 or -2*(c/k). I hope this will help you. See attached picture of the solution with arbitrary values of x between 0 and 1.

Dear Nicolus, I think you did a mistake: instead of sqrt(4C-4kEx^2) should be there sqrt(4C^2+4kEx^2).

Yes Lukáš Glod, thank you very much. I will go ahead and correct. In deed the matlab file was wrong too. So I hope it is corrected now. 

Thank you all above for your valuable suggestions. Sorry I have made slight correction to the equation above because the value of 2 should have multiplied both C and y values but this correction does not change the nature of the equation or the suggestions you forwarded. If t the algebraic solution is too complicated then i better use numerical methods because the end does not justify the means.

Dear Nicolus,

I can not quite figure out  how you came to z= y/x from  the original assumption z = dy/dx. Can you explain to me in more detail because this means that y/x = dy/dx = z which is not always the case for arbitrary functions and is true only for y = bx. However, and incidentally  the graph you produced looks pretty much like the one i obtained from MATLAB!

Substitution z= 2C-x, p=d(2c-y)/dx and x(p), z(p). The Legendre transformtion

z=a+x*p, a - intercept, p - slope of the tangent to the plot z(x), x=-da/dp, z=a+px=a-a'*p.

The equation is rearranged as

: Ep^2*x-zp-Ex=0 =>z/x=E*p-E/p=>-a/a'+p=Ep-E/p

Elementary integration a'/a=p/(p^2(1-E)+E)=> lna=const+(1/*(1-2E))ln(p^2*(1-E)+E)=>

a=K*((p^2*(1-E)+E))^(1/*(1-2E))=.x=-(...)', y=(...)-p*(,,,)'

ICs:x =0=>p=0 or +- infinity.

O.K

. but the solution is not sole and for any K=-inf...+inf incl K=0.

Very interesting equation

Thx

Error Z=2C-y. About succesfull use of the Legendre transform for solving non-linear ODEs see my papers on integration projectile motion in resistive medium. Everyone may download them

And what else,, The case E=1/2, Integral curve y=const. This is bifurcation point,

Dear Victor:

I can not thank you enough for your suggestion. Correct me if i am wrong.  I think the initial definition for z should be z = 2c-y not 2c-x (sorry you have already pointed out that i saw it later). Also the coefficient of integration in the expression lna=const+(1/*(1-2E))ln(p^2*(1-E)+E) should be (1/2) (1/(1-E)) not 1/(1-2E). According to your suggestion for the initial condition x=0 i should solve  x = -da/dp = 0. by differentiating the expression a=K*((p^2*(1-E)+E))^(1/*(1-2E)).  This gives a feasible point p=0 as you suggested. Then the value of a can be found from z= a+xp = 2C-y by substituting x=0, p=0 and y=o and it gives a = 2C for the initial condition. Finally i should use the expression a=K*((p^2*(1-E)+E))^(1/*(1-2E)) and evaluate the constant K by putting a=2C and P=0.  Many thanks again for teaching me this method.

Right You are. It was too late evening and I made several mistakes. If it isn't secret what do You need this equation for? Ecology? Biology?

I have derived this equation for the design of a shape of solar concentrator panel in order to dry/sterilize  fecal sludge material from urine diversion dry toilets. The parabolic solar panel concentrator is of course a special solution of the above equation which is y = mx^2 with E=1 and the value of C = 1/4m.

Why the shape differs from parabolic one? Because of extent focus domain for sludge? Is it possible to explain in details or give reference?

Yes it is because of the  focus domain.  For drying fecal sludge i thought areal focus is more appropriate than point focus because of the diffiulty of transmisson of heat within a larger size sludge media unllke for example water. The focus domain  is the sludge plate on which sludge is placed and which is a square /rectangle with smaller dimension value of 'a' and for a projected solar panel horizontal length of 'b' the value of E will be E= 1- 2a/b. In this way all the ray from the panel will concentrate on the plate dimension of a and as the sun moves this gives some flexibility. For a point focus as i said before a=0 and E=1 and the equation above will reduce to the parabolic mirror. The solar concentration ratio which can be taken as the total area of the panel divided by the area of the sludge plate can be determined from the solution of the equation above. The value of C in the above equation is, as i said before, the vertical distance from the ground for the sludge plate. By the way this is my imagination i would be happy to hear from you if it makes sense.

B. RADI, A. EL HAMI, Mathématique numérique pour l’ingénieur, Utilisation de MATLAB: Cours, exercices et problèmes de synthèse corrigés. Editions-Ellipses, Numéro ISBN 978-2-7298-5279-5, Novembre 2010 (240 pages, niveau second cycle)