Solution of the second equation is:

y(x)=g*(a-b*y)*(log(a-b*y)-1)+x*(g*log a)/b+(1-log a)/ga

Dear Dr

In fact I am not a Mathematician, do you have any links or books to demonstrate the solution you've found?

Thanks and sincerly regards

Dear Tahar,

   You can look for solution of similar equations in any elementary book on Differential Equations. In particular you can refer to Differential Equations by Simmons.

Thanks Akhlaq, I will

Tahar> I'm searching for the two diff. equations roots:

That sounds strange! Don't you search for solutions of the two differential equations? Roots is a concept for solutions of algebraic type of equations.

One may say that there is only one equation, since the first reduces to the second when k=0. The expression given by Akhlaq is not a correct solution. In the special case that (bg-ak)=0 there is a simple solution,

y(x)=k x2/(2b) ,

In the case that k=0 one may express x as function of y, but that expression involves a special function (the exponential integral). For k!=0 one may still express x as an integral involving y, but that integral cannot be expressed in terms of any named function.

From a practical point of view a numerical solution may be the best option. However, for qualitative understanding one can think of x as the time coordinate. Then the first equation can simply be interpreted as Newton's second law for a particle with position y, moving in a potential

V(y) = [(bg-ak)/b2] ln(a-b*y) - ky/b.

By plotting this potential for some relevant values of the parameters, and taking into account the initial conditions, you can learn a lot about the qualitative behaviour of the solution. Which can be very useful in combination with numerical integration.

Thanks Dr

But y shall be function of t (time) and of course the constants a, b and g, where would I find any reference of the equation you sent, and about the numerical integration, how could integrate such a function with three variable constants?

Tahar> But y shall be function of t (time)

Well, in your case the independent variable x plays the role of time. Take a look at the linked note about how to clean up the problem, and understand the qualitative behaviour of the solutions. Only after that does it make sense to solve the problem numerically (which otherwise most likely turns into a garbage in-garbage out process). Since you have a degree in Civil Engineering(?), you must have learned to solve elementary problems of mechanics.

Working Paper How to understand autonomous differential equations by just looking

Thanks Dr

I of course I will, I've a civil engineering degree, but I try to find a theoritical solutions for different problems in soil mechanics and in beams theory, such that: the copressibility of soils and the setelement of it under foundations, and to well understanding the real behaviour of beams, and I need of course the help of mathematics. I found myself really fulled of this domain of research: the applied mathematics in my speciality

Alternatively to what Kåre porposes. One may treat qualitative analysis as in the theory of dissipative systems. No integration in needed. Find attactors, analize them for stabily and so on. For example, the second equation may be written as

v'(x) = g/(a-b*y)

y'= v 

Hence

dv/dy = g/(a-b*y)/v

vdv = gdy/(a-b*y)

Kåre Olaussen Pr, would you send me your mobile number, I'd like to discuss with you about the project I indicated?

Best regards

ОК, that was a hint. Now I am going to finish.

vdv = gdy/(a-b*y) means

v^2/2 + g/b ln |y - a/b| = E

The fist term is kinetic energy, the second term is potential and the constant of intergration  is total energy. Never forget to take absolute value when the result of integration is logarithm.

Now it is easy to say what happens. 

Depending on the sign of g/b, the singularity is either up or down. 

When ti is down, the system would evolve to the point y = a/b moving ever faster. 

When it is up, the system would run away to infinity slowing down.

PROFIT!!!

For the first equation, an oscillatory regime seems possible since the potential energy contains a  linear term. But the analysis would require a piece of paper  I cannot determine whether (at what values of parameters) the initial point is in the correct half of the   phase space. You are welcome to finish.

Note: I am actually useful for simple problems  and Sergio Garcia has won the Masters.

This is the solution of the second equation

y(x)=g*(a-b*y)*(log(a-b*y)-1)+x*(g*log a)/b+(1-log a)/ga

You have an obvious typo Anna, since  y(0) = 0.

In that case

0 = ga(log a -1) + (1-log a)/ga

I don't think so Anna, because the exact solution I got contains inverse error function