Solution: The given equation is \begin{equation}\label{eq:04Apr2025-1} \left( 2x^2 + y \right) \ \mathrm{d} x + \left( x^2 y - x \right) \ \mathrm{d} y = 0. \end{equation} Let \[ M(x,y) = 2x^2 + y \quad \text{ and } \quad N(x,y) = x^2 y - x. \] Note that the equation is not exact as \[ \frac{\partial M}{\partial y} = 1 \quad \text{ and } \quad \frac{\partial N}{\partial x} = 2xy - 1 \] are not equal. Thus, we need to find an integrating factor. We have \begin{align*} \frac{\partial M}{\partial y} -\frac{\partial N}{\partial x} & = (1 - 2xy + 1) = 2 - 2xy. \end{align*} Note that, \begin{align*} \mu (x,y) & = \frac{\frac{\partial M}{\partial y} -\frac{\partial N}{\partial x}}{N} = \frac{2 - 2xy}{x^2 y - x} \\[1ex] & = \frac{2(1 - xy)}{x(xy - 1)} = -\frac{2}{x}. \end{align*} Thus, the integrating factor will be \begin{align*} e^{-\int \frac{2}{x}\ \mathrm{d} x} & = \frac{1}{x^2}. \end{align*}

Multiplying the given equation \eqref{eq:04Apr2025-1} by the integrating factor, we get \[ \frac{1}{x^2} \left( 2x^2 + y \right) \ \mathrm{d} x + \frac{1}{x^2} \left( x^2 y - x \right) \ \mathrm{d} y = 0. \] Let the solution be $u(x,y)$. Since the equation is exact, we have \begin{align*} \frac{\partial u}{\partial x} = \frac{M}{x^2} \quad \text{ and } \quad \frac{\partial u}{\partial y} = \frac{N}{x^2}. \end{align*} Thus, we have \begin{align*} \frac{\partial u}{\partial x} = 2x^2 + y \implies u(x,y) & = \int \left( 2 + \frac{y}{x^2} \right) \ \mathrm{d} x + g(y) \\ & = 2x - \frac{y}{x} + g(y). \end{align*} Now we also have \begin{align*} \frac{\partial u}{\partial y} = y - \frac{1}{x} & \implies - \frac{1}{x} + g'(y) = y - \frac{1}{x} \\ & \implies g'(y) = y \implies g(y) = \frac{y^2}{2}+ c. \end{align*} Thus, the solution will be \begin{align*} 2x - \frac{y}{x} + \frac{y^2}{2} + c = 0. \end{align*} We will now use the condition $y(1) = 2$ to find the constant $c$. \begin{align*} 2 - \frac{2}{1} + \frac{2^2}{2} + c = 0 \implies c = -2. \end{align*} Thus, the solution to the given initial value problem is \[ \textcolor{blue}{ \boxed{ 2x - \frac{y}{x} + \frac{y^2}{2} - 2 = 0. } } \]