function of x alone

 
 

Problem 02 | Determination of Integrating Factor

Problem 02
$2y(x^2 - y + x)~dx + (x^2 - 2y)~dy = 0$
 

Solution 02
$M~dx + N~dy = 0$

$2y(x^2 - y + x)~dx + (x^2 - 2y)~dy = 0$
 

$M = 2y(x^2 - y + x) = 2x^2y - 2y^2 + 2xy$

$N = x^2 - 2y$
 

$\dfrac{\partial M}{\partial y} = 2x^2 - 4y + 2x$

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

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = (2x^2 - 4y + 2x) - 2x$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = 2x^2 - 4y = 2(x^2 - 2y)$
 

Problem 01 | Determination of Integrating Factor

Problem 01
$(x^2 + y^2 + 1)~dx + x(x - 2y)~dy = 0$
 

Solution 01
$M~dx + N~dy = 0$

$(x^2 + y^2 + 1)~dx + x(x - 2y)~dy = 0$
 

$M = x^2 + y^2 + 1$

$N = x(x - 2y) = x^2 - 2xy$
 

$\dfrac{\partial M}{\partial y} = 2y$

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

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = 2y - (2x - 2y)$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = -2x + 4y$

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.

 

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