π((R+ΔR)2−R2)=(πΑR2+B)Δtwith R(0)=0π(2RΔR+ΔR2)=(πΑR2+B)ΔtdR=(CR+RD)dtwhereC=2A,D=2πBCR2+DRdR=dtln(CR2+D)=2C(t−G)CR2+D=e2CRt−GR=√Ce2C(t−G)−D=√CEe2Ct−Dfrom boundary conditionsR=√CD(e2Ct−1)=√D√Ce2Ct−1R=√2πB√2AeAt−1=√πAB√eAt−1Side note, limit when A→0matchequationsolvedforA=0.R′=√πAB⋅2√eAt−1eAt