Differential Equations: $(x - 2y - 1) dy = (2x - 4y - 5) dx$

By inspection, we can see that both sides of the equation contains x - 2y.
$(x - 2y - 1)\,dy = (2x - 4y - 5)\,dx$

$(z - 1)\,dy = (2z - 5)(dz + 2dy)$

$(z - 1)\,dy = (2z - 5)\,dz + 2(2z - 5)\,dy$

$(z - 1)\,dy - 2(2z - 5)\,dy = (2z - 5)\,dz$

$(z - 1 - 4z + 10)\,dy = (2z - 5)\,dz$

$(9 - 3z)\,dy = (2z - 5)\,dz$

$dy = \dfrac{(2z - 5)\,dz}{9 - 3z}$

$dy = \left( -\dfrac{2}{3} + \dfrac{1}{9 - 3z} \right) \, dz$

$y + c = -\frac{2}{3}z - \frac{1}{3}\ln (9 - 3z)$

$y + c = -\frac{2}{3}(x - 2y) - \frac{1}{3}\ln \Big[ 9 - 3(x - 2y) \Big]$

$-\frac{1}{3}y + c = -\frac{2}{3}x - \frac{1}{3}\ln (9 - 3x + 6y)$

$y + c = 2x + \ln (9 - 3x + 6y)$       answer