Problem 06 - 07 | Integrating Factors Found by Inspection

$y^3 \, dx + y \, dx + xy^2 \, dy - x \, dy = 0$

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

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

$(x \, dy + y \, dx) + \left( \dfrac{y \, dx - x \, dy}{y^2} \right) = 0$

$d(xy) + d\left( \dfrac{x}{y} \right) = 0$

$\displaystyle \int d(xy) + \int d\left( \dfrac{x}{y} \right) = 0$

$xy + \dfrac{x}{y} = c$

$xy^2 + x = cy$

$x(y^2 + 1) = cy$           answer

Divide by xy(y² + 1)
$\dfrac{y(y^2 + 1) \, dx}{xy(y^2 + 1)} + \dfrac{x(y^2 - 1) \, dy}{xy(y^2 + 1)} = 0$

$\dfrac{dx}{x} + \dfrac{(y^2 - 1) \, dy}{y(y^2 + 1)} = 0$
 

Resolve into partial fraction
$\dfrac{y^2 - 1}{y(y^2 + 1)} = \dfrac{A}{y} + \dfrac{By + C}{y^2 + 1}$

$y^2 - 1 = A(y^2 + 1) + (By + C)y$
 

Set y = 0, A = -1

Equate coefficients of y²
1 = A + B
1 = -1 + B
B = 2

Equate coefficients of y
0 = 0 + C
C = 0
 

Hence,
$\dfrac{y^2 - 1}{y(y^2 + 1)} = \dfrac{-1}{y} + \dfrac{2y}{y^2 + 1}$

$\dfrac{y^2 - 1}{y(y^2 + 1)} = \dfrac{2y}{y^2 + 1} - \dfrac{1}{y}$
 

Thus,
$\dfrac{dx}{x} + \left( \dfrac{2y}{y^2 + 1} - \dfrac{1}{y} \right) \, dy = 0$

$\displaystyle \int \dfrac{dx}{x} + \int \dfrac{2y \, dy}{y^2 + 1} - \int \dfrac{dy}{y} = 0$

$\ln x + \ln (y^2 + 1) - \ln y = \ln c$

$\ln \dfrac{x(y^2 + 1)}{y} = \ln c$

$\dfrac{x(y^2 + 1)}{y} = c$

$x(y^2 + 1) = cy$           answer - okay