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 $$p\left( t \right) = 400 - 300{e^{t/2}}$$