Official solution
Given differential equation is
\[{{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 200\]
By separating the variable, we get
\[dp\left( t \right) = \left[ {{1 \over 2}p\left( t \right) - 200} \right]dt\]
\[ \Rightarrow {{dp\left( t \right)} \over {{1 \over 2}p\left( t \right) - 200}} = dt\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\] Integrating on both the sides,
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {{{d\left( {p\left( t \right)} \right)} \over {{1 \over 2}p\left( t \right) - 200}}} = \int {dt} \]
Let \[{1 \over 2}p\left( t \right) - 200 = s \Rightarrow {{dp\left( t \right)} \over 2} = ds\]
So, \[\int {{{d\,p\left( t \right)} \over {\left( {{1 \over 2}p\left( t \right) - 200} \right)}}} = \int {dt} \]
\[ \Rightarrow \int {{{2ds} \over s} = \int {dt} } \Rightarrow 2\log s = t + c\]
\[ \Rightarrow 2\log \left( {{{p\left( t \right)} \over 2} - 200} \right) = t + c\]
\[ \Rightarrow {{p\left( t \right)} \over 2} - 200 = {e^{{1 \over 2}}}k\]
Using given condition \[p\left( t \right) = 400 - 300{e^{t/2}}\]