Final answer:
In the differential equation dy/dx + x³ + 2 = y/x, P(x) is 1/x and Q(x) is -x³ - 2. The integral of P with respect to x, ∫P dx, is the natural logarithm of the absolute value of x plus a constant of integration, ln|x| + C.
Step-by-step explanation:
The Ordinary Differential Equation given is dy/dx + x³ + 2 = y/x. To determine P and Q, we must first write the equation in the standard form of a linear first-order differential equation, which is dy/dx + P(x)y = Q(x). In this case, P(x) is 1/x and Q(x) is -x³ - 2. Once we have identified P and Q, we can find the integral ∫P dx, which is an integral of 1/x with respect to x.
Now, the integral ∫P dx = ∫(1/x) dx, which is a basic integral with a well-known solution. The solution to this integral is ln|x| + C, where C is the constant of integration.