general solution

Problem 02
$(x - 2y) \, dx + (2x + y) \, dy = 0$
 

Solution 02
$(x - 2y) \, dx + (2x + y) \, dy = 0$
 

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

Substitute,
$(x - 2vx) \, dx + (2x + vx)(v \, dx + x \, dv) = 0$

$x \, dx - 2vx \, dx + 2vx \, dx + 2x^2 \, dv + v^2x \, dx + vx^2 \, dv = 0$

$x \, dx + 2x^2 \, dv + v^2x \, dx + vx^2 \, dv = 0$

$(x \, dx + v^2x \, dx) + (2x^2 \, dv + vx^2 \, dv) = 0$

$x(1 + v^2) \, dx + x^2(2 + v) \, dv = 0$

Problem 01
$3(3x^2 + y^2) \, dx - 2xy \, dy = 0$
 

Solution 01
$3(3x^2 + y^2) \, dx - 2xy \, dy = 0$
 

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

Substitute,
$3(3x^2 + v^2x^2) \, dx - 2vx^2 (v \, dx + x \, dv) = 0$

$3(3 + v^2)x^2 \, dx - 2vx^2 (v \, dx + x \, dv) = 0$
 

Divide by x²,
$3(3 + v^2) \, dx - 2v (v \, dx + x \, dv) = 0$

$9 \, dx + 3v^2 \, dx - 2v^2 \, dx - 2vx \, dv = 0$

$9 \, dx + v^2 \, dx - 2vx \, dv = 0$

$(9 + v^2) \, dx - 2vx \, dv = 0$

Problem 22
$(xy + x) \, dx = (x^2y^2 + x^2 + y^2 + 1) \, dy = 0$
 

Solution 22
$(xy + x) \, dx = (x^2y^2 + x^2 + y^2 + 1) \, dy = 0$

$x(y + 1) \, dx = (x^2 + 1)(y^2 + 1) \, dy = 0$

$\dfrac{x(y + 1) \, dx}{(x^2 + 1)(y + 1)} = \dfrac{(x^2 + 1)(y^2 + 1) \, dy}{(x^2 + 1)(y + 1)} = 0$

$\dfrac{x \, dx}{x^2 + 1} = \dfrac{(y^2 + 1) \, dy}{y + 1} = 0$
 

By long division
$\dfrac{y^2 + 1}{y + 1} = y - 1 + \dfrac{2}{y + 1}$
 

Thus,
$\dfrac{x \, dx}{x^2 + 1} = \left( y - 1 + \dfrac{2}{y + 1} \right) \, dy = 0$

Problem 21
$x^2 \, dx + y(x - 1) \, dy = 0$
 

Solution 21
$x^2 \, dx + y(x - 1) \, dy = 0$

$\dfrac{x^2 \, dx}{x - 1} + y \, dy = 0$
 

By long division
$\dfrac{x^2 \, dx}{x - 1} = x + 1 + \dfrac{1}{x - 1}$
 

Thus,
$\left[ (x + 1) + \dfrac{1}{x - 1} \right] \, dx + y \, dy = 0$

$\displaystyle \int (x + 1) \, dx + \int \dfrac{dx}{x - 1} + \int y \, dy = 0$

$\frac{1}{2}(x + 1)^2 + \ln |x - 1| + \frac{1}{2}y^2 + \ln c = 0$

$(x + 1)^2 + 2\ln |x - 1| + y^2 + 2\ln c = 0$

Problem 20
$xy \, dx - (x + 2) \, dy = 0$
 

Solution 20
$xy \, dx - (x + 2) \, dy = 0$

$\dfrac{xy \, dx}{y(x + 2)} - \dfrac{(x + 2) \, dy}{y(x + 2)} = 0$

$\dfrac{x \, dx}{x + 2} - \dfrac{dy}{y} = 0$

$\left( 1 - \dfrac{2}{x + 2} \right) \, dx - \dfrac{dy}{y} = 0$

$\displaystyle \int dx - 2 \int \dfrac{dx}{x + 2} - \int \dfrac{dy}{y} = 0$

$x - 2 \ln (x + 2) - \ln y = \ln c$

$x = \ln c + \ln y + 2 \ln (x + 2)$

$x = \ln c + \ln y + \ln (x + 2)^2$

$x = \ln cy(x + 2)^2$

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

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

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

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

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

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

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

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

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

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