Problem 03
$(2xy - 3x^2) \, dx + (x^2 + y) \, dy = 0$
 

Solution 03
$(2xy - 3x^2) \, dx + (x^2 + y) \, dy = 0$
 

$M = 2xy - 3x^2$

$N = x^2 + y$
 

Test for exactness
$\dfrac{\partial M}{\partial y} = 2x$

$\dfrac{\partial N}{\partial x} = 2x$

Exact!
 

Let
$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = 2xy - 3x^2$

$\partial F = (2xy - 3x^2) \, \partial x$
 

Integrate partially in x, holding y as constant

Problem 02
$(6x + y^2) \, dx + y(2x - 3y) \, dy = 0$
 

Solution 02
$(6x + y^2) \, dx + y(2x - 3y) \, dy = 0$
 

$M = 6x + y^2$

$N = y(2x - 3y) = 2xy - 3y^2$
 

Test for exactness
$\dfrac{\partial M}{\partial y} = 2y$

$\dfrac{\partial N}{\partial x} = 2y$

Exact!
 

Let
$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = 6x + y^2$

$\partial F = (6x + y^2) \, \partial x$
 

Integrate partially in x, holding y as constant

Problem 01
$(x + y) \, dx + (x - y) \, dy = 0$
 

Solution 01
$(x + y) \, dx + (x - y) \, dy = 0$
 

Test for exactness
$M = x + y$   ;   $\dfrac{\partial M}{\partial y} = 1$

$N = x - y$   ;   $\dfrac{\partial N}{\partial x} = 1$

$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$   ;   thus, exact!
 

Step 1: Let
$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = x + y$
 

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

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

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

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

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