July 2011

Problem 04
$xy \, dx - (x^2 + 3y^2) \, dy = 0$
 

Solution 04
$xy \, dx - (x^2 + 3y^2) \, dy = 0$
 

Let
$x = vy$

$dx = v \, dy + y \, dv$
 

$vy^2(v \, dy + y \, dv) - (v^2y^2 + 3y^2) \, dy = 0$

$vy^2(v \, dy + y \, dv) - y^2(v^2 + 3) \, dy = 0$

$v(v \, dy + y \, dv) - (v^2 + 3) \, dy = 0$

$v^2 \, dy + vy \, dv - v^2 \, dy - 3 \, dy = 0$

$vy \, dv - 3 \, dy = 0$

$v \, dv - \dfrac{3 \, dy}{y} = 0$

$\displaystyle \int v \, dv - 3 \int \dfrac{dy}{y} = 0$

Problem 03
$2(2x^2 + y^2) \, dx - xy \, dy = 0$
 

Solution 03
$2(2x^2 + y^2) \, dx - xy \, dy = 0$
 

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

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

$4x^2 \, dx + 2v^2x^2 \, dx - v^2x^2 \, dx - vx^3 \, dv = 0$

$4x^2 \, dx + v^2x^2 \, dx - vx^3 \, dv = 0$

$x^2(4 + v^2) \, dx - vx^3 \, dv = 0$

$\dfrac{x^2(4 + v^2) \, dx}{x^3(4 + v^2)} - \dfrac{vx^3 \, dv}{x^3(4 + v^2)} = 0$

$\dfrac{dx}{x} - \dfrac{v \, dv}{4 + v^2} = 0$

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

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

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

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