Reduce order and respect a singular boundary

$z^{\prime}=(2x-1)y^{\prime\prime}+2y^{\prime}$. The given equation rearranges to $(2x-1)y^{\prime\prime}+2y^{\prime}=2$, so $dz/dx=2$.

$z=2x+C$, so $(2x-1)y^{\prime}=2x+C$. At $x=0$, $y^{\prime}=2$ gives $C=-2$, hence $y^{\prime}=(2x-2)/(2x-1)=1-1/(2x-1)$. Integration gives $y=x-\tfrac12\ln|2x-1|+B$; $y(0)=2$ gives $B=2$.

The initial point lies on $x<1/2$. Its maximal connected solution is $y=x-\tfrac12\ln(1-2x)+2$ on $(-\infty,1/2)$. The formula with the absolute value can also be drawn for $x>1/2$ using the same constant, as the source does, but this is a separate formal branch. The divergent curve cannot be continued through $x=1/2$, and the initial conditions do not fix an independent integration constant on the other side.

For the displayed two-branch formula, $y^{\prime}\to1$ as $x\to\pm\infty$; $y^{\prime}\to+\infty$ as $x\to(1/2)^-$ and $y^{\prime}\to-\infty$ as $x\to(1/2)^+$. Only the negative-infinity and left-hand singular limits belong to the connected initial-value solution. A source line in (iii) prints $1-1/[2(2x-1)]$, inserting an extra factor of two; the earlier derivation and the derivative of the stated solution give $1-1/(2x-1)$. The reported limit signs remain correct.

The initial-value branch passes through $(0,2)$ and increases to $+\infty$ at the vertical asymptote $x=1/2$. The independently displayed right branch decreases from $+\infty$, reaches $(1,3)$, and then increases. The sketch uses a solid initial-value branch and a dashed formal right branch. A limiting gradient of $1$ does not establish an oblique line asymptote: the logarithmic term continues to grow in magnitude.