Answer
$$
\frac{dP}{dt} =\sqrt {Pt}, \quad P(1)=2 .
$$
The solution of the given differential equation is
$$
P=\left[\frac{1}{3}t^{\frac{3}{2}}+(\sqrt {2}- \frac{1}{3})\right] ^{2}
$$
Work Step by Step
$$
\frac{dP}{dt} =\sqrt {Pt}, \quad P(1)=2 .
$$
We write the equation in terms of differentials:
$$
\frac{dP}{\sqrt {P}} =\sqrt {t} dt
$$
and integrate both sides:
$$
\int \frac{dP}{\sqrt {P}}=\int\sqrt {t} dt
$$
$ \Rightarrow$
$$
2P^{\frac{1}{2}}=\frac{2}{3}t^{\frac{3}{2}}+C
$$
where $C $ is an arbitrary constant.
Now, to satisfy the condition $ P(1)=2, $we have
$$
2(2)^{\frac{1}{2}}=\frac{2}{3}+C \Rightarrow C =2\sqrt {2}- \frac{2}{3}
$$
So, the solution of the given differential equation is
$$
2P^{\frac{1}{2}}=\frac{2}{3}t^{\frac{3}{2}}+(2\sqrt {2}- \frac{2}{3})
$$
In other words, it can be written in the form:
$$
P=\left[\frac{1}{3}t^{\frac{3}{2}}+(\sqrt {2}- \frac{1}{3})\right] ^{2}
$$
