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Solve the foregoing problem if lambda_(1)=lambda_(2)=lambda |
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Answer» Solution :(a) This case can be OBTAINED from the previous one on putting `lambda_(2)=lambda_(2)- epsilon` where `epsilon` is very SMALL and letting `epsilon rarr 0` at the end. Then `N_(2)=(lambda_(1)N_(10))/(epsilon)(e^(epsilont)-1)e^(-lambda_(1)t)= lambda_(1)te^(-lambda_(1)t)N_(10)` or dropping the subscrtipt 1 as the two VALUES are equal `N_(2)=N_(10)lambdate^(-lambda t)` (b) This is maximum when `(dN_(2))/(DT)=0 or t=(1)/(lambda)` |
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