Differential Equations: $(r - 3s - 7) dr = (2r - 4s - 10) ds$

Coefficient Linear in Two Variables
$(r- 3s - 7) \, dr = (2r - 4s - 10) \, ds$

$\Big[ (x + 1) - 3(y - 2) - 7 \Big] \, dx = \Big[ 2(x + 1) - 4(y - 2) - 10 \Big] \, dy$

$(x + 1 - 3y + 6 - 7) \, dx = (2x + 2 - 4y + 8 - 10) \, dy$

$(x - 3y) \, dx = (2x - 4y) \, dy$   ←   homogeneous

$(x - 3vx) \, dx = (2x - 4vx)(v\,dx + x\,dv)$

$(x - 3vx) \, dx = v(2x - 4vx)\,dx + x(2x - 4vx)\,dv$

$\Big[ (x - 3vx) - v(2x - 4vx) \Big]\,dx = x(2x - 4vx)\,dv$

$(x - 3vx - 2vx + 4v^2x)\,dx = x(2x - 4vx)\,dv$

$(x - 5vx + 4v^2x)\,dx = x(2x - 4vx)\,dv$

$x(1 - 5v + 4v^2)\,dx = x^2(2 - 4v)\,dv$

$\dfrac{x\,dx}{x^2} = \dfrac{(2 - 4v)\,dv}{1 - 5v + 4v^2}$

$\dfrac{dx}{x} = \dfrac{(2 - 4v)\,dv}{(v - 1)(4v - 1)}$

$\dfrac{dx}{x} = \left( \dfrac{-2/3}{v - 1} + \dfrac{-4/3}{4v - 1} \right) \, dv$

$\displaystyle \int \dfrac{dx}{x} = -\dfrac{2}{3} \int \dfrac{dv}{v - 1} - \dfrac{1}{3} \int \dfrac{4\,dv}{4v - 1}$

$\ln x + c = -\frac{2}{3}\ln (v - 1) - \frac{1}{3}\ln (4v - 1)$

$\ln x + c = -\frac{1}{3} \Big[ 2\ln (v - 1) + \ln (4v - 1) \Big]$

$\ln x + c = -\frac{1}{3} \Big[ \ln (v - 1)^2 + \ln (4v - 1) \Big]$

$\ln x + c = -\frac{1}{3} \ln \Big[ (v - 1)^2(4v - 1) \Big] $

$\ln x + c = -\frac{1}{3} \ln \left[ \left( \dfrac{y}{x} - 1 \right)^2 \left( \dfrac{4y}{x} - 1 \right) \right]$

$\ln x + c = -\frac{1}{3} \ln \left[ \left( \dfrac{y - 1}{x}\right)^2 \left( \dfrac{4y - 1}{x}\right) \right]$

$\ln x + c = -\frac{1}{3} \ln \dfrac{(y - 1)^2(4y - 1)}{x^3}$

$\ln (r - 1) + c = -\frac{1}{3} \ln \dfrac{[ (s + 2) - 1)^2 [ 4(s + 2) - 1 ]}{(r - 1)^3}$

$\ln (r - 1) + c = -\frac{1}{3} \ln \dfrac{(s + 1)^2(4s + 7)}{(r - 1)^3}$

Note: double check everything