Find the solution of the differential equation that satisfies ${{dP} \over {dt}} = 8\sqrt {Pt} ,\,P(1) = 5$ Please help. My homework is grading my answer as incorrect, but I can't tell what I did wrong. The second photo is the work of the problem done correctly but with dp/dt=2sqrt(Pt). I based my work off of this, but my answer is incorrect. Thank you!
The link to the question is:
Q3
and here is my answer:
C

$$\frac{dP}{dt}=8\sqrt{Pt}$$
$$\frac{dP}{\sqrt{P}}=8\sqrt{t}dt$$
$$\int \frac{dP}{\sqrt{P}}=\int 8\sqrt{t}dt$$
$$2\sqrt{P}=8\cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c=\frac{16}{3}t^{\frac{3}{2}}+c$$
By initial conditions, $P(1)=5$
$$2\sqrt{5}=\frac{16}{3}+c$$
$$c=2\sqrt{5}-\frac{16}{3}$$
The answer should be $$2\sqrt{P}=\frac{16}{3}t^{\frac{3}{2}}+2\sqrt{5}-\frac{16}{3}$$
$$\sqrt{P}=\frac{8}{3}t^{\frac{3}{2}}+\sqrt{5}-\frac{8}{3}$$
$$P=\left(\frac{8}{3}t^{\frac{3}{2}}+\sqrt{5}-\frac{8}{3}\right)^2$$
