Answer
$\sqrt P = \frac{{{t^{3/2}}}}{3} + \sqrt 2 - \frac{1}{3}$
Work Step by Step
$$\eqalign{
& \frac{{dP}}{{dt}} = \sqrt {Pt} ,{\text{ }}P\left( 1 \right) = 2 \cr
& {\text{Using the radical property }}\sqrt {ab} = \sqrt a \sqrt b \cr
& \frac{{dP}}{{dt}} = \sqrt P \sqrt t \cr
& {\text{Separating the variables}} \cr
& \frac{{dP}}{{\sqrt P }} = \sqrt t dt \cr
& {P^{ - 1/2}}dP = {t^{1/2}}dt \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {{P^{ - 1/2}}dP} = \int {{t^{1/2}}dt} dx \cr
& \frac{{{P^{1/2}}}}{{1/2}} = \frac{{{t^{3/2}}}}{{3/2}} + C \cr
& 2\sqrt P = \frac{{2{t^{3/2}}}}{3} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}P\left( 1 \right) = 2 \cr
& 2\sqrt 2 = \frac{{2{{\left( 1 \right)}^{3/2}}}}{3} + C \cr
& 2\sqrt 2 - \frac{2}{3} = C \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& 2\sqrt P = \frac{{2{t^{3/2}}}}{3} + 2\sqrt 2 - \frac{2}{3} \cr
& \sqrt P = \frac{{{t^{3/2}}}}{3} + \sqrt 2 - \frac{1}{3} \cr} $$
