To determine whether the given differential equation is exact, we need to check if it satisfies the condition:
∂M/∂y = ∂N/∂x
where M and N are the coefficients of dy and dx, respectively, in the differential equation:
(2sin(x)cos(y) − 2y^2)dy - (xsin^2(y) - 4x^2)dx = 0
Now, let's find the partial derivatives:
∂M/∂y = 2cos(x)cos(y) - 4y
∂N/∂x = -sin^2(y) - 8x
The equation is exact if ∂M/∂y = ∂N/∂x. Let's check:
2cos(x)cos(y) - 4y = -sin^2(y) - 8x
Rearranging the equation, we get:
sin^2(y) - 2cos(x)cos(y) - 8x + 4y = 0
Since ∂M/∂y = ∂N/∂x, the given differential equation is exact.
To solve the exact differential equation, we need to find a potential function φ(x, y) such that:
∂φ/∂x = M
∂φ/∂y = N
Integrating M with respect to x, we get:
φ(x, y) = ∫(2sin(x)cos(y) - 2y^2)dx
= 2∫sin(x)cos(y)dx - 2∫y^2dx
= -2cos(x)cos(y) - 2xy^2 + g(y)
Here, g(y) is the constant of integration with respect to x.
Next, we differentiate φ(x, y) partially with respect to y and equate it to N:
∂φ/∂y = N
-2sin(x)sin(y) - 4xy + g'(y) = -sin^2(y) - 8x
From this equation, we can equate the coefficients of the terms involving y:
-2sin(x)sin(y) - 4xy = -8x
-2sin(x)sin(y) - 4xy + 8x = 0
Comparing the coefficients, we have:
g'(y) = 0
Integrating g'(y), we obtain:
g(y) = c
Here, c is the constant of integration with respect to y.
Finally, the potential function φ(x, y) becomes:
φ(x, y) = -2cos(x)cos(y) - 2xy^2 + c
Therefore, the solution to the given exact differential equation is φ(x, y) = -2cos(x)cos(y) - 2xy^2 + c, where c is an arbitrary constant.