Loading Documentπ((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
Method 2
x′=Ax+BSolve related:x′=Axx=CeAtBack with C(x):AC(x)eAt+C′(x)eAt=AC(x)eAt+BC′(x)=Be−AtC(x)=A−Be−At+KSo:x=KeAt−ABAnd from x(0)x=AB(eAt−1)EventuallyπR2=AB(eAt−1)R =√AπB√(eAt−1)