Question

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

Answer 1

$\displaystyle(4xy + 2x^2) \mathrm{d}x + (2x^2+3y^2) \mathrm{d}y=0\\ M = (4xy + 2x^2), N = (2x^2+3y^2)\\ M_y = \frac{\partial M}{\partial y} = 4x\\ N_x = \frac{\partial N}{\partial x} = 4x\\ M_y = N_x, \\ \textsf{so the differential equation is exact.}\\ f(x, y) = \int 4xy + 2x^2\, \mathrm{d}x\\ f(x, y) = 2x^2y + \frac{2x^3}{3} + f(y) \\ \begin{aligned} f(x, y) &= \int 2x^2+3y^2\, \mathrm{d}y \\&= 2x^2y + y^3 + f(x) \end{aligned}\\ \therefore f(x, y) = 2x^2y + \frac{2x^3}{3} + y^3 \\ \textsf{is the solution to the exact}\\\textsf{differential equation}$