# Differential equation with partial sum as coefficient [closed]

I saw the following differential equation/solution in a paper and it is puzzling me. Any help with regards to the nature of this differential equation (with partial sum as coefficient?) or the specifics of the derivation are highly appreciated!

.... Y(t) must satisfy the differential equation

## -$\frac{d}{dt} Y(t) = a^{-1}\left[1+\sum_{i=1}^{N}b_i\right]Y(t)$

with the initial condition Y(0)=1. Integration yields

## closed as unclear what you're asking by zhoraster, user91500, TastyRomeo, Namaste, Daniel W. FarlowFeb 2 '17 at 13:25

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The equation is just $$Y't()=A Y(t)$$ which gives $$Y(t)=C e^{At}$$ and the condition gives $C=1$. So $$Y(t)=\exp\left(-\frac{1+\sum_{i=1}^n b_i} a t\right)$$ which write $$\log(Y(t))=-\frac t a-\frac{\sum_{i=1}^n b_i} a t=-\frac t a-\frac{b_1} a t-\frac{b_2} a t-\frac{b_3} a t+\cdots$$ $$Y(t)=\exp(-\frac t a)\exp(-\frac{b_1} a t)\exp(-\frac{b_2} a t)\exp(-\frac{b_3} a t)\cdots\exp(-\frac{b_n} a t)$$ whic is your formula.