Question

Find the solution of the differential equation dP/dt=2(sqrt(Pt)), that satisfies the given initial condition P(1)=2

Answer 1

Is that
\[\frac{dP}{dt}=2\sqrt{Pt}~~?\]

Answer 2

If it is, then the equation is separable:
\[\frac{dP}{dt}=2\sqrt{Pt}~~\iff~~\frac{dP}{\sqrt P}=2\sqrt t~dt\]

Answer 3

that is correct. Where do i go from there?
Thanks for the help!

Answer 4

Recall that \(\sqrt x=x^{1/2}\) and \(\dfrac{1}{\sqrt x}=\dfrac{1}{x^{1/2}}=x^{-1/2}\). It's power rule from there.

Answer 5

\[P^{-1/2}~dP=2t^{1/2}~dt\]
Integrating both sides yields
\[\frac{P^{1/2}}{1/2}=2\frac{t^{3/2}}{3/2}+C~~\iff~~2P^{1/2}=\frac{4}{3}t^{3/2}+C\]

Answer 6

Solve for \(C\) with the given condition, \(P(1)=2\):
\[2(2^{1/2})=\frac{4}{3}(1^{3/2})+C~~\Rightarrow~~2^{3/2}=\frac{4}{3}+C~~\Rightarrow~~C=\cdots\]
then plug it into the general solution to get the particular one.

Answer 7

so is the answer [P=\frac{ 1 }{ 2 }(\frac{ 4 }{ 3 }t ^{3/2}+\sqrt{3}/2)^{2}\]

Answer 8

\[\begin{align*}2P^{1/2}&=\frac{4}{3}t^{3/2}+\underbrace{2^{3/2}-\frac{4}{3}}_{C}\\\\
P^{1/2}&=\frac{2}{3}t^{3/2}+2^{1/2}-\frac{2}{3}\\\\
P&=\left(\frac{2}{3}t^{3/2}+\sqrt2-\frac{2}{3}\right)^2
\end{align*}\]
I'd stop there, but you could expand if you like