Final answer:
To find the limit as x approaches 0 for the given initial-value problem, we can solve the differential equation and substitute the initial condition. However, the limit does not exist because the natural logarithm of 0 is undefined.
Step-by-step explanation:
To find the limit as x approaches 0, we need to find the value of y(x) when x is very close to 0. We can do this by solving the given initial-value problem. First, let's find the general solution of the differential equation x²y' - xy = -16xcos(4x). We can rewrite the equation as y' - (1/x)y = -16cos(4x). This is a first-order linear homogeneous equation, which can be solved using an integrating factor.
Multiplying both sides of the equation by the integrating factor, which is e-ln(x) = 1/x, we get (1/x)y' - (1/x²)y = -16cos(4x)/x. Integrating both sides with respect to x gives us ln|x|y = C - 16sin(4x). Rearranging the equation gives us y = (C - 16sin(4x))/ln|x|.
To find the particular solution that satisfies the initial condition y(π) = 0, we substitute x = π and y = 0 into the equation. This gives us 0 = (C - 16sin(4π))/ln|π|. Since ln|π| is a constant, the numerator must be 0. Therefore, C = 16sin(4π). Substituting this value back into the general solution gives us y = (16sin(4π) - 16sin(4x))/ln|x|.
Taking the limit as x approaches 0, we get y = (16sin(4π) - 16sin(0))/ln|0|. However, ln|0| is undefined, so the limit does not exist. Therefore, the answer is A. The limit does not exist.