Answer
$${\text{true}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{100}}{{1 + 99{e^{ - 5t}}}} \cr
& {\text{write as }}y = 100{\left( {1 + 99{e^{ - 5t}}} \right)^{ - 1}} \cr
& y = 100{\left( {1 + 99{e^{ - 5t}}} \right)^{ - 1}} \cr
& {\text{differentiate with respet to }}t \cr
& \frac{{dy}}{{dt}} = 100\frac{d}{{dt}}{\left( {1 + 99{e^{ - 5t}}} \right)^{ - 1}} \cr
& \frac{{dy}}{{dt}} = 100\left( { - 1} \right){\left( {1 + 99{e^{ - 5t}}} \right)^{ - 2}}\left( {99} \right)\left( { - 5{e^{ - 5t}}} \right) \cr
& \frac{{dy}}{{dt}} = \frac{{49,500{e^{ - 5t}}}}{{{{\left( {1 + 99{e^{ - 5t}}} \right)}^2}}} \cr
& \frac{{dy}}{{dt}} = \left( {\frac{{495{e^{ - 5t}}}}{{1 + 99{e^{ - 5t}}}}} \right)\left( {\frac{{100}}{{1 + 99{e^{ - 5t}}}}} \right) \cr
& \frac{{dy}}{{dt}} = \left( {\frac{{495{e^{ - 5t}}}}{{{{\left( {1 + 99{e^{ - 5t}}} \right)}^2}}}} \right)y \cr
& {\text{we have the differetial equation }}\frac{{dy}}{{dt}} = 5\left( {1 - \frac{y}{{100}}} \right)y{\text{ then }} \cr
& \frac{{495{e^{ - 5t}}}}{{{{\left( {1 + 99{e^{ - 5t}}} \right)}^2}}} = 5\left( {1 - \frac{y}{{100}}} \right) \cr
& {\text{replacing }}y = \frac{{100}}{{1 + 99{e^{ - 5t}}}} \cr
& \frac{{495{e^{ - 5t}}}}{{{{\left( {1 + 99{e^{ - 5t}}} \right)}^2}}} = 5\left( {1 - \frac{1}{{100}}\left( {\frac{{100}}{{1 + 99{e^{ - 5t}}}}} \right)} \right) \cr
& {\text{simplifying}} \cr
& \frac{{495{e^{ - 5t}}}}{{{{\left( {1 + 99{e^{ - 5t}}} \right)}^2}}} = 5\left( {1 - \frac{1}{{1 + 99{e^{ - 5t}}}}} \right) \cr
& \frac{{495{e^{ - 5t}}}}{{{{\left( {1 + 99{e^{ - 5t}}} \right)}^2}}} = 5\left( {\frac{{1 + 99{e^{ - 5t}} - 1}}{{1 + 99{e^{ - 5t}}}}} \right) \cr
& \frac{{495{e^{ - 5t}}}}{{{{\left( {1 + 99{e^{ - 5t}}} \right)}^2}}} = 5\left( {\frac{{99{e^{ - 5t}}}}{{1 + 99{e^{ - 5t}}}}} \right) \cr
& \frac{{495{e^{ - 5t}}}}{{{{\left( {1 + 99{e^{ - 5t}}} \right)}^2}}} = \frac{{495{e^{ - 5t}}}}{{{{\left( {1 + 99{e^{ - 5t}}} \right)}^2}}} \cr
& {\text{then we can say that }}y = \frac{{100}}{{1 + 99{e^{ - 5t}}}}{\text{ satifies the differential equation}} \cr
& {\text{true}} \cr} $$
