A NonLinear Differential Equation.

Let $z = \dfrac{dy}{dx} \implies \dfrac{dz}{dx} = \dfrac{d^2y}{dx^2} \implies$ $\dfrac{dz}{dx} + z = z^2 \implies \dfrac{dz}{dx} = z(z - 1) \implies$

$\displaystyle\int \dfrac{dz}{z(z - 1)} = \displaystyle\int dx$

$\displaystyle\int \dfrac{dz}{z(z - 1)} = \displaystyle\int (\dfrac{1}{z - 1} - \dfrac{1}{z}) dz =$ $\ln(\dfrac{z - 1}{z}) = x + C \implies \dfrac{z - 1}{z} = c_{1}e^{x} \implies$

$z = \dfrac{1}{1 - c_{1}e^x} \implies \dfrac{dy}{dx} = \dfrac{1}{1 - c_{1}e^{x}} \implies$ $y = \displaystyle\int \dfrac{dx}{1 - c_{1}e^x}$

Let $u = e^x \implies du = e^x dx \implies \dfrac{1}{u} du = dx \implies$

$\displaystyle\int \dfrac{dx}{1 - c_{1}e^x} = \dfrac{1}{u(1 - c_{1}u)} du =$

$\dfrac{1}{u(1 - c_{1}u)} \dfrac{a}{u} + \dfrac{B}{1 - c_{1}u} \implies$ $(B - c_{1}A)u + A = 1 \implies B - c_{1}A = 0$ and

$A = 1 \implies B = c_{1} \implies$ $y = \displaystyle\int (\dfrac{1}{u} + \dfrac{c_{1}}{1 - c_{1}u}) du =$ $\ln(u) - \ln(1 - c_{1}) + c_{2} =$

$\ln(\dfrac{u}{1 - c_{1}u}) + c_{2} = \ln(\dfrac{e^x}{1 - c_{1}e^x}) + c_{2}$

$\therefore y(x) = \ln(\dfrac{e^x}{1 - c_{1}e^x}) + c_{2}$ .

Using the Initial conditions are $y(0) = 0$ and $y(\ln(3)) = \ln(2)$ we have:

$y(0) = 0 = \ln(\dfrac{1}{1 - c_{1}}) + c_{2} \implies c_{2} = \ln(1 - c_{1})$

and

$y(\ln(3)) = \ln(2) = \ln(\dfrac{3}{1 - 3c_{1}}) + \ln(1 - c_{1}) = \ln(\dfrac{3 - 3c_{1}}{1 - 3c_{1}}) \implies$

$2 = \dfrac{3 - 3c_{1}}{1 - 3c_{1}} \implies 2 - 6c_{1} = 3 - 3c_{1} \implies c_{1} = -\dfrac{1}{3} \implies$

$c_{2} = \ln(\dfrac{4}{3}) \implies y_{p}(x) = \ln(\dfrac{4e^x}{3 + e^x})$

Checking we obtain:

$\dfrac{dy_{p}}{dx} = \dfrac{3}{3 + e^x}$ and $\dfrac{d^2y_{p}}{dx^2} = \dfrac{-3e^x}{(3 + e^x)^2}$ and $(\dfrac{dy_{p}}{dx})^2 = \dfrac{9}{(3 + e^x)^2}$

Substituting into the initial differential we have:

$\dfrac{-3e^x}{(3 + e^x)^2} + \dfrac{3}{3 + e^x} = \dfrac{9}{(3 + e^x)^2} = \dfrac{d^2y_{p}}{dx^2}$ .

$\lim_{x \rightarrow +\infty} y_{p}(x) = \lim_{x \rightarrow +\infty} \ln(\dfrac{4e^x}{3 + e^x}) =$ $\ln(\lim_{x \rightarrow +\infty} \dfrac{4e^x}{3 + e^x}) = \ln(4) \approx \boxed{1.386294}$