If y = y(x) is the solution of the differential equation 2 x ^ 2...

Given $2{x}^{2}\dfrac{dy}{dx} - 2xy + 3{y}^{2} = 0$

Dividing by $2{x}^{2}{y}^{2}$, we get,

$- \dfrac{1}{{y}^{2}}\dfrac{dy}{dx} + \dfrac{1}{xy} = \dfrac{3}{2{x}^{2}}$

Let $\dfrac{1}{y} = p, \dfrac{- 1}{{y}^{2}}\dfrac{dy}{dx} = \dfrac{dp}{dx}$

$⇒ \dfrac{dp}{dx} + \dfrac{1}{x}\left(p\right) = \dfrac{3}{2{x}^{2}}$

General solution will be $p \cdot {e}^{∫ \dfrac{1}{x}dx} = ∫ \dfrac{3}{2{x}^{2}}{\cdot}^{{e}^{∫ \dfrac{1}{x}dx}}dx + c$
$⇒ \dfrac{x}{y} = ∫ \dfrac{3}{2x}dx + c ⇒ \dfrac{x}{y} = \dfrac{3}{2}\ln x + c$

$∵ y\left(e\right) = \dfrac{e}{3} ⇒ c = \dfrac{3}{2} ⇒ \dfrac{x}{y} = \dfrac{3}{2}\ln x + \dfrac{3}{2}$ is the particular solution.

when $x = 1; y\left(1\right) = \dfrac{2}{3}$