Problem 19
$dr = b(\cos \theta \, dr + r \sin \theta \, d\theta)$
 

Solution 19
$dr = b(\cos \theta \, dr + r \sin \theta \, d\theta)$

$dr = b\cos \theta \, dr + br \sin \theta \, d\theta$

$dr - b\cos \theta \, dr = br \sin \theta \, d\theta$

$(1 - b\cos \theta) \, dr = br \sin \theta \, d\theta$

$\dfrac{(1 - b\cos \theta) \, dr}{r(1 - b\cos \theta)} = \dfrac{br \sin \theta \, d\theta}{r(1 - b\cos \theta)}$

$\dfrac{dr}{r} = \dfrac{b \sin \theta \, d\theta}{1 - b\cos \theta}$

Problem 18
$ye^{2x} \, dx = (4 + e^{2x}) \, dy$
 

Solution 18
$ye^{2x} \, dx = (4 + e^{2x}) \, dy$

$\dfrac{ye^{2x} \, dx}{y(4 + e^{2x})} = \dfrac{(4 + e^{2x}) \, dy}{y(4 + e^{2x})}$

$\dfrac{e^{2x} \, dx}{4 + e^{2x}} = \dfrac{dy}{y}$

$\displaystyle \dfrac{1}{2} \int \dfrac{e^{2x} (2 \, dx)}{4 + e^{2x}} = \int \dfrac{dy}{y}$

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

$\frac{1}{2} \ln (4 + e^{2x}) = \ln cy$

$\ln (4 + e^{2x}) = 2\ln cy$

$\ln (4 + e^{2x}) = \ln (cy)^2$

$\ln (4 + e^{2x}) = \ln c^2y^2$

Problem 14
$2y \, dx = 3x \, dy$

Solution 14

Problem 15
$my \, dx = nx \, dy$

Solution 15
$my \, dx = nx \, dy$

$m\dfrac{dx}{x} = n\dfrac{dy}{y}$

$m\ln x = n\ln y + \ln c$

$\ln x^m = \ln y^n + \ln c$

$\ln x^m = \ln cy^n$

Problem 12
$\sin x \sin y \, dx + \cos x \cos y \, dy = 0$
 

Solution 12
$\sin x \sin y \, dx + \cos x \cos y \, dy = 0$

$\dfrac{\sin x \sin y \, dx}{\sin y \cos x} + \dfrac{\cos x \cos y \, dy}{\sin y \cos x} = 0$

$\dfrac{\sin x \, dx}{\cos x} + \dfrac{\cos y \, dy}{\sin y} = 0$

$\displaystyle -\int \dfrac{-\sin x \, dx}{\cos x} + \int \dfrac{\cos y \, dy}{\sin y} = 0$

$-\ln (\cos x) + \ln (\sin y) = \ln c$

$\ln \dfrac{\sin y}{\cos x}= \ln c$

$\dfrac{\sin y}{\cos x}= c$