Current ramps up, and yes, the diode will be forward biased. At t < 0, the diode's junction voltage is undefined (you didn't give initial conditions); if we assume zero initial voltage and current, then: the junction voltage will start at zero, and charge up as the inductor begins to draw current. Later, the diode voltage saturates to Vf (the DC value, given as a function of current), as it goes into conduction proper.
If the inductor and diode are ideal, and Vf << (5V supply), then the diode voltage will be a pretty nice logarithm, and the current will be a linear ramp. (Well, okay, for an ideal diode -- with no junction capacitance -- the diode forward voltage will appear immediately, without taking time to charge the capacitance.)
In a real circuit, with resistance, the current will saturate, according to the L/R time constant, and the diode voltage as well. As \$t \rightarrow \infty\$, the system reaches steady state.
The voltage drop across the inductor, at t = 0, will be exactly 5V, because Vf(If = 0) = 0. The derivative of Vf and If depends on capacitance: if capacitance is zero, the derivatives are undefined at t = 0. If nonzero, all derivatives exist, and the first derivative is zero. I think?
Clear as mud?
Tim