Solve the following differential equation (7xy2−3)dx+(7x2y+4)dy=0

Answer 1

M(x,y) = 7xy² - 3; N(x, y) = 7x²y + 4; M_y = N_x = 14xy, so this is exact equation.

F_x = 7xy² - 3; F(x, y) = ∫(7xy² - 3)dx = 7/2x²y² - 3x + f(y); F_y = 7x²y + f'(y) = 7x²y + 4; f'(y) = 4; f(y) = 4y + C.

F(x, y) = 7/2x²y² - 3x + 4y + C