Problem 02 | Equations with Homogeneous Coefficients

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

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

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

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

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

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

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

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

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

$\displaystyle \int \dfrac{dx}{x} + \int \left( \dfrac{2}{1 + v^2} + \dfrac{v}{1 + v^2} \right) \, dv = 0$

$\displaystyle \int \dfrac{dx}{x} + 2 \int \dfrac{dv}{1 + v^2} + \dfrac{1}{2}\int \dfrac{2v \, dv}{1 + v^2} = 0$

$\ln x + 2 \arctan v + \frac{1}{2} \ln (1 + v^2) = \frac{1}{2}c$

$2 \ln x + 4 \arctan v + \ln (1 + v^2) = c$

$\ln x^2 + 4 \arctan v + \ln (1 + v^2) = c$

$\ln [ \, x^2(1 + v^2) \, ] + 4 \arctan v = c$
 

From
$y = vx$

$v = \dfrac{y}{x}$
 

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

$\ln (x^2 + y^2) + 4 \arctan (y/x) = c$           answer