Answer
$$\frac{y}{{1 - y/10}} = \frac{{70}}{3}{e^t}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = 2.8y\left( {1 - y/10} \right) \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{{y\left( {1 - y/10} \right)}} = dt \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{{dy}}{{y\left( {1 - y/10} \right)}}} = \int {dt} \cr
& {\text{Decompose into partial fractions }}\frac{1}{{y\left( {1 - y/10} \right)}} \cr
& \frac{1}{{y\left( {1 - y/10} \right)}} = \frac{A}{y} + \frac{B}{{1 - y/10}} \cr
& 1 = A\left( {1 - \frac{y}{{10}}} \right) + By \cr
& y = 0 \to A = 1 \cr
& y = 10 \to B = \frac{1}{{10}} \cr
& \frac{1}{{y\left( {1 - y/10} \right)}} = \frac{1}{y} + \frac{1}{{10\left( {1 - y/10} \right)}} \cr
& {\text{Then,}} \cr
& \int {\left[ {\frac{1}{y} + \frac{1}{{10\left( {1 - y/10} \right)}}} \right]} dy = \int {dt} \cr
& {\text{Integrating both sides}} \cr
& {\text{ln}}\left| y \right| - \ln \left| {1 - y/10} \right| = t + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Using the initial condition }}\left( {0,7} \right) \cr
& {\text{ln}}\left| 7 \right| - \ln \left| {1 - 7/10} \right| = 0 + C \cr
& \ln 7 - \ln 0.3 = C \cr
& C = \ln \left( {\frac{{70}}{3}} \right) \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& {\text{ln}}\left| y \right| - \ln \left| {1 - y/10} \right| = t + \ln \left( {\frac{{70}}{3}} \right) \cr
& {\text{ln}}\left| {\frac{y}{{1 - y/10}}} \right| = t + \ln \left( {\frac{{70}}{3}} \right) \cr
& \frac{y}{{1 - y/10}} = \frac{{70}}{3}{e^t} \cr} $$
