I'm just wondering why you'd bother making a substitution when the equation is first order linear?
$\displaystyle \begin{align*} x\,\frac{dy}{dx} - y - 2x^2 + 1 &= 0 \\ x\,\frac{dy}{dx} - y &= 2x^2 - 1 \\ \frac{dy}{dx} - \frac{1}{x}\,y &= 2x - \frac{1}{x} \end{align*}$
Now the integrating factor is $\displaystyle \begin{align*} e^{\int{-\frac{1}{x}\,dx}} = e^{-\ln{(x)}} = e^{\ln{ \left( x^{-1} \right) } } = x^{-1} \end{align*}$, multiplying through we get
$\displaystyle \begin{align*} x^{-1} \left( \frac{dy}{dx} - x^{-1}\,y\right) &= x^{-1} \left( 2x - x^{-1} \right) \\ x^{-1}\,\frac{dy}{dx} - x^{-2}\,y &= 2 - x^{-2} \\ \frac{d}{dx} \left( x^{-1}\,y \right) &= 2 - x^{-2} \\ x^{-1}\,y &= \int{2 - x^{-2}\,dx} \\ x^{-1}\,y &= 2x + x^{-1} + C \\ y &= 2x^2 + 1 + C\,x \end{align*}$
From here it can be seen that $\displaystyle \begin{align*} \frac{dy}{dx} = 4x + C \end{align*}$, and since there is a stationary point on the positive x-axis, that means there's a point x where both y = 0 and $\displaystyle \begin{align*} \frac{dy}{dx} = 0 \end{align*}$
So $\displaystyle \begin{align*} 4x + C = 0 \end{align*}$ and $\displaystyle \begin{align*} 2x^2 + 1 + C\,x = 0 \end{align*}$. From the first of these, we find $\displaystyle \begin{align*} C = -4x \end{align*}$. Substituting into the second gives
$\displaystyle \begin{align*} 2x^2 + 1 - 4x^2 &= 0 \\ 1 - 2x^2 &= 0 \\ 1 &= 2x^2 \\ \frac{1}{2} &= x^2 \\ \pm \frac{\sqrt{2}}{2} &= x \end{align*}$
Since we are told x is positive, that means we set $\displaystyle \begin{align*} x = \frac{\sqrt{2}}{2} \end{align*}$, and since $\displaystyle \begin{align*} C = -4x \end{align*}$, that means $\displaystyle \begin{align*} C = -4 \left( \frac{\sqrt{2}}{2} \right) = -2\sqrt{2} \end{align*}$.
Thus the particular solution to this DE is $\displaystyle \begin{align*} y = 2x^2 - 2\sqrt{2}\,x + 1 \end{align*}$.