Final answer:
The given differential equation (2xy^2 − 7) dx + (2x^2y + 6) dy = 0 is exact. To solve it, we need to find a function F(x,y) such that the partial derivatives of F with respect to x and y are equal to the given coefficients of the equation. By integrating the partial derivatives and comparing coefficients, we can find the solution: x^2y^2 - 7x + 6y + c = 0.
Step-by-step explanation:
A differential equation is said to be exact if it can be written in the form M(x,y) dx + N(x,y) dy = 0, where M and N are functions of x and y whose partial derivatives with respect to y and x, respectively, are equal: ∂M/∂y = ∂N/∂x.
In the given differential equation, (2xy^2 − 7) dx + (2x^2y + 6) dy = 0, the equation is exact because ∂M/∂y = 4xy = ∂N/∂x. To solve the exact differential equation, we need to find a function F(x,y) such that ∂F/∂x = M and ∂F/∂y = N. By integrating the partial derivatives of M and N with respect to x and y, we can find F.
Integrating the partial derivative of M, we get F = ∫(2xy^2 − 7) dx = x^2y^2 - 7x + g(y), where g(y) is a function of y. Now, we differentiate this expression with respect to y and equate it to the N expression to find g(y). We have ∂F/∂y = ∂/∂y (x^2y^2 - 7x + g(y)) = 2xy^2 + g'(y) = N = 2x^2y + 6.
Comparing the coefficients, we get g'(y) = 6, so g(y) = 6y + c, where c is a constant. Substituting g(y) back into the expression for F, we get F = x^2y^2 - 7x + 6y + c. Thus, the solution to the exact differential equation is x^2y^2 - 7x + 6y + c = 0, where c is a constant.