Nonlinear ODE of second order with Boundary Conditions defined. The problem is:
$y''(x)-a\cdot(y(x))^2=0, a>0$
S.t. $ y(0)=b,
\lim_{x\to\infty } y'(x)=0$
That problem results from a catalyst which has a chemical reaction of second order occuring within it - the book Transport Phenomena of Bird at. al. contains that question. Someone can give me a tip how to proceed to solve that nonlinear ODE? That is the first time that I found this kind of problem.
 A: $$y''-a\:y^2=0\quad\to\quad 2y''y'-2a\:y^2y'=0$$
$$y'^2-\frac{2a}{3}y^3=c_1$$
$$y'=\pm \sqrt{c_1+\frac{2a}{3}y^3}$$
$$dx=\pm\frac{dy}{\sqrt{c_1+\frac{2a}{3}y^3}}\quad\to\quad x=\pm\int \frac{dy}{\sqrt{c_1+\frac{2a}{3}y^3}}$$
This integral involves the elliptic functions, which would be rather arduous.
By luck, we will see later than the specified boundary conditions are not sufficient, which means that they are an infinity of solutions. If we want to easily find not all solutions, but only one solution, we can take the liberty of arbitrary choosing $c_1$, for example $c_1=0$. This implicitly supposes that $\quad y(x\to\infty)\to 0.\quad$ In this particular case :
$$x=\pm\sqrt{\frac{3}{2a}}\int \frac{dy}{y^{3/2}}=\pm\sqrt{\frac{6}{a\:y}}+c_2$$
$$y=\frac{6}{a(x-c_2)^2}$$
$$y'=-\frac{12}{a(x-c_2)^3}$$
We see that the condition $\lim_{x\to\infty } y'(x)=0$ is satisfied.
Condition $\quad y(0)=b=\frac{6}{a(-c_2)^2} \quad\to\quad c_2=\pm \sqrt{\frac{6}{ab}}. \quad$ The sign minus is selected in order to avoid a discontinuity for $y(x)$ at $x=\sqrt{\frac{6}{ab}}$
$$y(x)=\frac{6}{a\left(x+\sqrt{\frac{6}{ab}}\right)^2}$$
We see that a solution which satisfies the ODE and the specified boudbary conditions is obtained even if the constant $c_1$ was arbitrary chosen. This proves that the conditions specified in the wording of the problem are insufficient to define an unique solution. The above solution is only one among the infinity of solutions. But it is probably the simplest.
A: You start by setting $y'=z$ in order to come up with a first order separable differential equation. 
(Sorry for the short answer but I thought since you asked for a tip I should not write the whole solution).
