Solve Cauchy-Euler boundary value problem $4x^2y'' + 8xy' -3y=0$ subject to initial conditions $y(0) = 0\, y(4) = 6$

Question

I have to solve $4x^2y'' + 8xy' -3y=0$ subject to initial conditions $y(0) = 0,\, y(4) = 6$

This is a Cauchy-Euler equation so using the characteristic equation, I get the general solution $y = Ax^{-\frac{3}{2}} + Bx^{\frac{1}{2}}$

However, the first initial condition seems nonsense as $y(0) = 0 \implies$ $A(0) + B(0) = 0$, which leaves:

$$y(4) = \frac{1}{8}A + 2B = 6$$

Which is not enough information to give the particular solution. Is there something else I should be using?

Answer 1

As $0^{-3/2}$ does not exist, your initial conditions imply $A=0$ and $B=3$.