Answer to Question #146805 in Differential Equations for Jd Lumbania

Question

(2xy^2 - 2y)dx + (3x^2y - 4x)dy = 0

Answer 1

$$Solution$$

$\frac{∂}{∂y}(2xy^2 - 2y) - \frac{∂}{∂x}(3x^2y - 4x) = (4xy - 2) - (6xy - 4) = -2xy + 2.$

Since $\frac{[\frac{∂}{∂y}(2xy^2 - 2y) - \frac{∂}{∂x}(3x^2y - 4x)] }{ (2xy^2 - 2y)} =\frac{ -2(xy - 1)}{2y(xy - 1) }= \frac{-1}{y}$ , a function of y alone,

we know that $e^{∫ (\frac{1}{y}) dy} = y$ is an integrating factor.

Multiply both sides of the DE by $y$ :

$(2xy^3 - 2y^2) dx + (3x^2y^2 - 4xy) dy = 0.$

Now, this is an exact DE, because

$\frac{∂}{∂y}(2xy^3 - 2y^2) = 6xy^2 - 4y = \frac{∂}{∂x}(3x^2y^2 - 4xy).$

Since

$∫ (2xy^3 - 2y^2) dx = x^2 y^3 - 2xy^2 + f(y)\\ ∫ (3x^2y^2 - 4xy) dy = x^2 y^3 - 2xy^2 + g(x),$

the general solution is $x^2 y^3 - 2xy^2 = C$ .