To solve this differential equation, we first need to check if it is exact. To do this, we compute the partial derivatives of the terms involving x and y:
∂/∂y (х^2 + y) = 1 ∂/∂x (x - 2y) = 1
Since these partial derivatives are equal, the differential equation is exact.
Next, we find the potential function F(x, y) such that:
∂F/∂x = х^2 + y ∂F/∂y = x - 2y
Integrating the first equation with respect to x gives:
F = (1/3) * x^3 + xy + g(y)
Differentiating this result with respect to y and comparing with the second equation gives:
∂F/∂y = x + g'(y) = x - 2y
Therefore, g'(y) = -2y, which means g(y) = -y^2.
So the potential function F(x, y) is:
F = (1/3) * x^3 + xy - y^2
Therefore, the general solution of the differential equation is:
To solve this differential equation, we first need to check if it is exact. To do this, we compute the partial derivatives of the terms involving x and y:
∂/∂y (х^2 + y) = 1
∂/∂x (x - 2y) = 1
Since these partial derivatives are equal, the differential equation is exact.
Next, we find the potential function F(x, y) such that:
∂F/∂x = х^2 + y
∂F/∂y = x - 2y
Integrating the first equation with respect to x gives:
F = (1/3) * x^3 + xy + g(y)
Differentiating this result with respect to y and comparing with the second equation gives:
∂F/∂y = x + g'(y) = x - 2y
Therefore, g'(y) = -2y, which means g(y) = -y^2.
So the potential function F(x, y) is:
F = (1/3) * x^3 + xy - y^2
Therefore, the general solution of the differential equation is:
(1/3) * x^3 + xy - y^2 = C
where C is the constant of integration.