How do you solve for v_o in this clipper circuit?

Here is the circuit:I tried to solve it, but I am just not sure if my solution is correct.First I assumed that d2 is off for the positive half part of the cycle, then the circuit is simplified to this:

d1 is on when vo is larger than 2 (vd=0) so:$$2=v_{i}\frac{2}{3}$$$$v_{i}=3$$From there, I applied nodal analysis to solve for vo when vi >3$$ v_{i}-v_{o}=\frac{v_{o}}{2}+v_{o}-2$$Is this the way to go, or am I far from the right answer?