NonLinear Differential.

Let $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} = (\dfrac{dy}{dx})^3$

Let $z = \dfrac{dy}{dx} \implies \dfrac{dz}{dx} = \dfrac{d^2y}{dx^2} \implies$ $\dfrac{dz}{dx} = z^3 - z = z(z - 1)(z + 1) \implies$

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

$\dfrac{1}{z(z - 1)(z + 1)} = \dfrac{A}{z} + \dfrac{B}{z - 1} + \dfrac{C}{z + 1} \implies$

$A(z^2 - 1) + B(z^2 + z) + C(z^2 - z) = 1 \implies A + B + C = 0, B - C = 0$ and

$A = -1 \implies B = \dfrac{1}{2} = C \implies$

$\displaystyle\int -\dfrac{1}{z} dz + \dfrac{1}{2}\displaystyle\int (\dfrac{1}{z - 1} + \dfrac{1}{z + 1}) dz$ $= \ln(z) + \dfrac{1}{2}(\ln(z - 1) + \ln(z + 1)) =$

$\ln(\dfrac{\sqrt{z^2 - 1}}{z}) = x + C \implies \dfrac{\sqrt{z^2 - 1}}{z} = c_{1}e^{2x} \implies$

$z^2 - 1 = z^2(c_{1}^2 e^{2x}) \implies z^2(1 - c_{1}^2 e^{2x}) = 1$ $\implies z = \dfrac{1}{\sqrt{1 - c_{1}^2 e^{2x}}} \implies$

$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - c_{1}^2 e^{2x}}} \implies y = \displaystyle\int \dfrac{dx}{\sqrt{1 - c_{1}^2 e^{2x}}}$

Let $c_{1}e^x = \sin(\theta) \implies c_{1}e^x dx = \sin(\theta) dx = \cos(\theta) d\theta \implies$ $dx = \dfrac{\cos(\theta)}{\sin(\theta)} d\theta \implies$

$y = \displaystyle\int \csc(\theta) d\theta = \ln(|\csc(\theta) - \cot(\theta)|) =$ $\ln(|\dfrac{1 - \sqrt{1 - c_{1}^2 e^{2x}}}{c_{1}e^x}|) + c_{2}$

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

$y(0) = 0$ and $y(\ln(2)) = \ln(\dfrac{1}{2 - \sqrt{3}})$

$y_{0} = 0 \implies c_{2} = \ln(|\dfrac{c_{1}}{1 - \sqrt{1 - c_{1}^2}}|)$

$\implies y = \ln(|\dfrac{1 - \sqrt{1 - c_{1}^2 e^{2x}}}{e^x(1 - \sqrt{1 - c_{1}^2})}|)$

$y(\ln(2)) = \ln(\dfrac{1}{2 - \sqrt{3}}) \implies$ $\dfrac{1}{2 - \sqrt{3}} = \dfrac{1 - \sqrt{1 - c_{1}^2 e^{2x}}}{e^x(1 - \sqrt{1 - c_{1}^2})} \implies$

$2(1 - \sqrt{1 - c_{1}^2}) = (2 - \sqrt{3})(1 - \sqrt{1 - 4c_{1}^2}) \implies$

$-2\sqrt{1 - c_{1}^2} = -\sqrt{3} - (2 - \sqrt{3})\sqrt{1 - 4c_{1}^2} \implies$

$(2 - \sqrt{3})\sqrt{1 - 4c_{1}^2} = 2\sqrt{1 - c_{1}^2} - \sqrt{3} \implies$

$(2 - \sqrt{3})^2(1 - 4c_{1}^2) = 7 - 4c_{1}^2 - 4\sqrt{3}\sqrt{1 - c_{1}^2} \implies$

$7 - 4c_{1}^2 - (2 - \sqrt{3})^2 + 4(2 - \sqrt{3})^2c_{1}^2 = 4\sqrt{3}\sqrt{1 - c_{1}^2} \implies$

$(4(7 - 4\sqrt{3}) - 4)c_{1}^2 + 7 - (7 - 4\sqrt{3}) = 4\sqrt{3}\sqrt{1 - c_{1}^2} \implies$

$8(3 - 2\sqrt{3})c_{1}^2 + 4\sqrt{3} = 4\sqrt{3}\sqrt{1 - c_{1}^2} \implies$

$4\sqrt{3}(2(\sqrt{3} - 2)c_{1}^2 + 1) = 4\sqrt{3}\sqrt{1 - c_{1}^2} \implies$

$2(\sqrt{3} - 2)c_{1}^2 + 1 = \sqrt{1 - c_{1}^2}$

$4(\sqrt{3} - 2)^2c_{1}^4 + 4(\sqrt{3} - 2)c_{1}^2 + 1 = 1 - c_{1}^2 \implies$

$4(7 - 4\sqrt{3})c_{1}^4 - (7 - 4\sqrt{3})c_{1}^2 = 0 \implies$

$(7 - 4\sqrt{3})c_{1}^2(4c_{1}^2 - 1) = 0$ and $c_{1} \neq 0 \implies c_{1} = \pm\dfrac{1}{2}$

$\implies y_{p}(x) = \ln(|\dfrac{2 - \sqrt{4 - e^{2x}}}{(2 - \sqrt{3})e^x}|)$

We don't need the absolute value here so that $\boxed{y_{p}(x) = \ln(\dfrac{2 - \sqrt{4 - e^{2x}}}{(2 - \sqrt{3})e^x})}$ , where the

domain $D = \{x \in \mathbb{R} \mid x \leq \ln(2)\}$ and the range $R = \{y \in \mathbb{R} \mid y \leq \ln(\dfrac{1}{2 - \sqrt{3}})\}$ .

Checking $y_{p}(x)$ :

$\dfrac{dy_{p}}{dx} = \dfrac{2}{(4 - e^{2x})^{\frac{1}{2}}}$ $\implies$ $\dfrac{d^2y_{p}}{dx^2} = \dfrac{2e^{2x}}{(4 - e^{2x})^{\frac{3}{2}}}$

$\implies$

$\dfrac{d^2y_{p}}{dx^2} + \dfrac{dy_{p}}{dx} = \dfrac{8}{(4 - e^{2x})^{\frac{3}{2}}} = (\dfrac{2}{\sqrt{4 - e^{2x}}})^{3} =$ $(\dfrac{dy_{p}}{dx})^3$

and

$y_{p}(\dfrac{3}{2}) = \ln(\dfrac{4 - \sqrt{7}}{3(2 - \sqrt{3})}) =$ $\ln(\dfrac{2^2 - \sqrt{7}}{3(2 - \sqrt{3})}) =$ $\ln(\dfrac{a^a - \sqrt{b}}{c(a - \sqrt{c})})$

$\implies a + b + c = \boxed{12}$ .