Final answer:
W= 4xy'' + y' = 0 is a subspace of the vector space of all functions because it is closed under addition, scalar multiplication, and contains the zero vector. The closure is verified by showing that any linear combination of members of W satisfies the defining differential equation.
Step-by-step explanation:
To determine if the set W= 4xy'' + y' = 0 is a subspace of the vector space of all functions, we must verify if W is closed under addition and scalar multiplication and contains the zero vector. Let's consider two functions y1(x) and y2(x) that belong to W, meaning that they both satisfy the differential equation 4xy'' + y' = 0. If we take their sum, y1(x) + y2(x), and differentiate it, the resulting linear combination should also satisfy the differential equation, which we can check by substituting into the differential equation:
(4x(y1(x) + y2(x))'' + (y1(x) + y2(x))') = (4xy1''(x) + y1'(x)) + (4xy2''(x) + y2'(x)) = 0 + 0 = 0,
which demonstrates closure under addition. For a scalar c, c·y(x) should also satisfy the differential equation, meaning that 4x(cy)'' + (cy)' = c(4xy'' + y') = 0, indicating closure under scalar multiplication. Finally, the zero function y(x) = 0 satisfies the differential equation, hence the zero vector is in W.