Problem 04 | Determination of Integrating Factor

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Problem 04
$y(4x + y)~dx - 2(x^2 - y)~dy = 0$

Solution 04

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$M~dx + N~dy = 0$

$y(4x + y)~dx - 2(x^2 - y)~dy = 0$

$M = y(4x + y) = 4xy + y^2$

$N = -2(x^2 - y) = -2x^2 + 2y$

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

$\dfrac{\partial N}{\partial x} = -4x$

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

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = 8x + 2y$

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

$\dfrac{1}{N}\left( \dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} \right) = \dfrac{2(4x + y)}{-2(x^2 - y)}$ → neither a function of x alone nor y alone

$\dfrac{1}{M}\left( \dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} \right) = \dfrac{2(4x + y)}{y(4x + y)} = \dfrac{2}{y}$ → a function y alone