Capacitor discharging into another capacitor with RC timing

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cupid.callin
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1. The problem statement and my attempt
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Its not the correct answer.
answer is: Q/2(1 - e-2t/RC)
 

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Use conservation of charge; there's nowhere for it to escape to!
 
The sum of the voltages across R and C2 is equal to the voltage of the first capacitor. Q1/C1 =I*R +Q2/C2. As the charge is conserved, Q1+Q2=Qo=const, and I=dQ2/dt=-dQ1/dt, so you have the differential equation (Qo-Q2)C1+Q2/C2+RdQ2/dt=0 with the initial condition Q2(0)=0.

ehild