I am going to assume that there is a typo in the statement of the problem otherwise the boundary condition at zero isn't needed as the solution to the homogenous problem is

Which is bounded no matter what the constants are but if we solve the problem

Now this is in its self adjoint form.

Solving the homogeneous equation gives

Now we need to find two solutions each satisfying one of the boundary conditions.

other wise the log function is unbounded at zero and

to give zero at 1.

So the Green's function has the form

Now if we take the Wronskian we get

Now we take the function off of the derivative of y in the self adjoint form and multiply it by the Wronskian and set it equal to negative 1 to get

So we can pick any real numbers for A and B that satisfy this equation so let

So the Green's function is

So the solution to the ODE is

Notice this satisfies the boundary conditions and that

and

If you plug this back into the ODE you get