The topic is Additional Topics in Ordinary DE of the first order.
(r-3s-7)dr=(2r-4s-10)ds
Coefficient Linear in Two Variables $(r- 3s - 7) \, dr = (2r - 4s - 10) \, ds$
From (1) and (2) r = 1 and s = -2
Let r = x + 1 → dr = dx s = y - 2 → ds = dy
$\Big[ (x + 1) - 3(y - 2) - 7 \Big] \, dx = \Big[ 2(x + 1) - 4(y - 2) - 10 \Big] \, dy$
$(x + 1 - 3y + 6 - 7) \, dx = (2x + 2 - 4y + 8 - 10) \, dy$
$(x - 3y) \, dx = (2x - 4y) \, dy$ ← homogeneous
$(x - 3vx) \, dx = (2x - 4vx)(v\,dx + x\,dv)$
$(x - 3vx) \, dx = v(2x - 4vx)\,dx + x(2x - 4vx)\,dv$
$\Big[ (x - 3vx) - v(2x - 4vx) \Big]\,dx = x(2x - 4vx)\,dv$
$(x - 3vx - 2vx + 4v^2x)\,dx = x(2x - 4vx)\,dv$
$(x - 5vx + 4v^2x)\,dx = x(2x - 4vx)\,dv$
$x(1 - 5v + 4v^2)\,dx = x^2(2 - 4v)\,dv$
$\dfrac{x\,dx}{x^2} = \dfrac{(2 - 4v)\,dv}{1 - 5v + 4v^2}$
$\dfrac{dx}{x} = \dfrac{(2 - 4v)\,dv}{(v - 1)(4v - 1)}$
$2 - 4v = A(4v - 1) + B(v - 1)$
When v = 1, A = -2/3 When v = 1/4, B = -4/3
$\dfrac{dx}{x} = \left( \dfrac{-2/3}{v - 1} + \dfrac{-4/3}{4v - 1} \right) \, dv$
$\displaystyle \int \dfrac{dx}{x} = -\dfrac{2}{3} \int \dfrac{dv}{v - 1} - \dfrac{1}{3} \int \dfrac{4\,dv}{4v - 1}$
$\ln x + c = -\frac{2}{3}\ln (v - 1) - \frac{1}{3}\ln (4v - 1)$
$\ln x + c = -\frac{1}{3} \Big[ 2\ln (v - 1) + \ln (4v - 1) \Big]$
$\ln x + c = -\frac{1}{3} \Big[ \ln (v - 1)^2 + \ln (4v - 1) \Big]$
$\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
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Coefficient Linear in Two Variables
$(r- 3s - 7) \, dr = (2r - 4s - 10) \, ds$
2r - 4s - 10 = 0 ← (2)
From (1) and (2)
r = 1 and
s = -2
Let
r = x + 1 → dr = dx
s = y - 2 → ds = dy
$\Big[ (x + 1) - 3(y - 2) - 7 \Big] \, dx = \Big[ 2(x + 1) - 4(y - 2) - 10 \Big] \, dy$
$(x + 1 - 3y + 6 - 7) \, dx = (2x + 2 - 4y + 8 - 10) \, dy$
$(x - 3y) \, dx = (2x - 4y) \, dy$ ← homogeneous
y = vx
dy = v dx + x dv
$(x - 3vx) \, dx = (2x - 4vx)(v\,dx + x\,dv)$
$(x - 3vx) \, dx = v(2x - 4vx)\,dx + x(2x - 4vx)\,dv$
$\Big[ (x - 3vx) - v(2x - 4vx) \Big]\,dx = x(2x - 4vx)\,dv$
$(x - 3vx - 2vx + 4v^2x)\,dx = x(2x - 4vx)\,dv$
$(x - 5vx + 4v^2x)\,dx = x(2x - 4vx)\,dv$
$x(1 - 5v + 4v^2)\,dx = x^2(2 - 4v)\,dv$
$\dfrac{x\,dx}{x^2} = \dfrac{(2 - 4v)\,dv}{1 - 5v + 4v^2}$
$\dfrac{dx}{x} = \dfrac{(2 - 4v)\,dv}{(v - 1)(4v - 1)}$
$\dfrac{2 - 4v}{(v - 1)(4v - 1)} = \dfrac{A}{v - 1} + \dfrac{B}{4v - 1}$
$2 - 4v = A(4v - 1) + B(v - 1)$
When v = 1, A = -2/3
When v = 1/4, B = -4/3
$\dfrac{dx}{x} = \left( \dfrac{-2/3}{v - 1} + \dfrac{-4/3}{4v - 1} \right) \, dv$
$\displaystyle \int \dfrac{dx}{x} = -\dfrac{2}{3} \int \dfrac{dv}{v - 1} - \dfrac{1}{3} \int \dfrac{4\,dv}{4v - 1}$
$\ln x + c = -\frac{2}{3}\ln (v - 1) - \frac{1}{3}\ln (4v - 1)$
$\ln x + c = -\frac{1}{3} \Big[ 2\ln (v - 1) + \ln (4v - 1) \Big]$
$\ln x + c = -\frac{1}{3} \Big[ \ln (v - 1)^2 + \ln (4v - 1) \Big]$
$\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
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