Final answer:
The differential equation (7x + 3y)dx + (3x - 8y³)dy = 0 is exact because the partial derivative of (7x + 3y) with respect to y is equal to the partial derivative of (3x - 8y³) with respect to x. The equation is solved by finding an integrating factor F(x, y) such that F(x, y) = 7/2 x² + 3xy - 2y´ = C, where C is a constant.
Step-by-step explanation:
To determine whether the given differential equation is exact, we need to check if the partial derivatives of the functions multiplied by dx and dy are equal. We have the differential equation (7x + 3y)dx + (3x - 8y³)dy = 0.
The function multiplied by dx is M(x, y) = 7x + 3y, and the partial derivative with respect to y is M_y = 3. The function multiplied by dy is N(x, y) = 3x - 8y³, and the partial derivative with respect to x is N_x = 3.
Since M_y = N_x, the differential equation is exact. To solve it, we look for a function F(x, y) such that F_x = M and F_y = N. Integrating M with respect to x gives F(x, y) = 7/2 x² + 3xy + g(y), where g(y) is an unknown function depending only on y.
Next, we differentiate this F(x, y) with respect to y and compare it to N. We get F_y = 3x + g'(y) = N = 3x - 8y³. Equating the terms not involving x, we find that g'(y) = -8y³. Integrating g'(y) with respect to y, we get g(y) = -2y´.
Thus, the solution to the differential equation is F(x, y) = 7/2 x² + 3xy - 2y´ = C, where C is a constant.