Question #496b1

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Answer 1 · 2017-08-14T20:16:48

$100 e^{- 0.2 t} .$ $100e^(-0.2t).$

Given that, $\frac{d N}{d t} = K N .$ $(dN)/dt=KN.$

$:. (dN)/N=Kdt,.....".[Separable Variable Type Diff. Eqn.]."$

Integrating, $int(dN)/N=Kintdt+C.$

$:. lnN=Kt+C.$

$:. N=e^(Kt+C)=e^(Kt)*e^C=ce^(Kt)," where, c=e^C."$

With, $c=100, &, K=-0.2," we have,"$

$N=100e^(-0.2t).$

Hence, the Last Option is the Right Choice.