A NonLinear Differential Equation !

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

$\implies \dfrac{d^2y}{dx^2} = (\dfrac{dy}{dx})^2 - \dfrac{1}{x}\dfrac{dy}{dx}$

Let $z = \dfrac{dy}{dx} \implies \dfrac{dz}{dx} = \dfrac{d^2y}{dx^2} \implies$

$\dfrac{dz}{dx} = z^2 - \dfrac{1}{x}z \implies \dfrac{dz}{dx} + \dfrac{1}{x}z = z^2 \implies$

$z^{-2}\dfrac{dz}{dx} + \dfrac{1}{x}z^{-1} = 1$

Let $w = z^{-1} \implies \dfrac{dw}{dx} = -z^{-2}\dfrac{dz}{dx} \implies$ $\dfrac{dz}{dx} = -z^2\dfrac{dw}{dx} \implies$

$\dfrac{dw}{dx} - \dfrac{1}{x}w = -1$ which is a first order linear differential equation $\implies$

$\dfrac{d}{dx}(e^{\displaystyle\int \dfrac{-1}{x}dx} * w) = - e^{\displaystyle\int \dfrac{-1}{x}dx} \implies$

$\dfrac{w}{x} = -\displaystyle\int \dfrac{1}{x} dx + c_{1} \implies w = x(-\ln(x) + c_{1})$

$w = \dfrac{1}{z} \implies z = \dfrac{1}{x(-\ln(x) + c_{1})}$ and $z = \dfrac{dy}{dx} \implies$

$y = -\displaystyle\int \dfrac{-\dfrac{1}{x}}{-\ln(x) + c_{1}} dx = -\ln(c_{1} - \ln(x)) + c_{2}$

$\boxed{y(x) = -\ln(c_{1} - \ln(x)) + c_{2}}$

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

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

$\implies c_{1} = 2\ln(2) = \ln(4)$

$\implies y_{p}(x) = \ln(\dfrac{\ln(4)}{\ln(4) - \ln(x)})$ which has a vertical asymptote at $x = 4 \implies \boxed{a = 4}$ .

Note: Using $y_{p}(x) = \ln(\dfrac{\ln(4)}{|\ln(4) - \ln(|x|)|}) \implies$ vertical asymptotes at $x = \pm 4$ .