July 2011

Problem 04
$(y^2 - 2xy + 6x) \, dx - (x^2 - 2xy + 2) \, dy = 0$
 

Solution 04
$(y^2 - 2xy + 6x) \, dx - (x^2 - 2xy + 2) \, dy = 0$

$M = y^2 - 2xy + 6x$

$N = -x^2 + 2xy - 2$
 

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

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

Exact!
 

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

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

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

Integrate partially in x, holding y as constant

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

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