mastevano,
You need to supply more detail on the inductive load. With a "perfect" inductor with zero DC resistance, the initial voltage across the inductor/cap combination will be zero. When you open the switch, the bulk of the initial inductor current will flow through the suppression diode, which will limit the voltage to -.7 volts. The uP input will be at zero volts at the instant you open the switch.
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Without the suppression diode, the intial uP input would still be zero, but the L/C circuit would exhibit a damped oscillation, starting out in the negative direction from zero.
Regards,
Kral