exact equation

 
 

The Determination of Integrating Factor

From the differential equation
 

$M ~ dx + N ~ dy = 0$

 

Rule 1
If   $\dfrac{1}{N}\left( \dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} \right) = f(x)$,   a function of x alone, then   $u = e^{\int f(x)~dx}$   is the integrating factor.

 

Rule 2
If   $\dfrac{1}{M}\left( \dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} \right) = f(y)$,   a function of y alone, then   $u = e^{-\int f(y)~dy}$   is the integrating factor.

 

Problem 04 | Exact Equations

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 | Exact Equations

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 | Exact Equations

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 | Exact Equations

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

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

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