Question

Find the solution of the differential equation dP/dt = sqrt(P*t) that satisfies the initial condition P(1)=5. Answer: P(t) = ....

Answer 1

\[\Large \frac{ dP }{ dt } = \sqrt{Pt}\]

Answer 2

\[\large \frac{ (dP)^2 }{ (dt)^2 } = Pt\]

Answer 3

\[\large \sqrt{\frac{ 1 }{ P }}dP = \sqrt{t}dt\]

Answer 4

integrate both sides and use initial conditions
\[\large 2(5)^{\frac{1}{2}} = \frac{2}{3}(1)^{\frac{3}{2}} + C\]

Answer 5

\[\large C = 2\sqrt{5} - \frac{2}{3}\]

Answer 6

So,
\[\large P(t) = {(\frac{2}{6}t^{\frac{2}{3}}+\sqrt{5} - \frac{2}{6})}^{2}\]

Answer 7

Answer is wrong...please help me find my mistake

Answer 8

@estudier

Answer 9

\[\frac{dP}{dt}=\sqrt{Pt} \rightarrow \frac{dP}{\sqrt{P}}=\sqrt{t}dt\]

Answer 10

rest like you did

Answer 11

I agree with your integral, so maybe it is the algebra, still checking......

Answer 12

@myko : that is what i did, just left out the step that included doing that

Answer 13

after integral i get\[\large 2P^{\frac{1}{2}} = \frac{2}{3}t^{\frac{3}{2}} + C\]

Answer 14

that's same what i got, now just put initial condition

Answer 15

which i sub in t=1 and P=5 and i get: 2sqrt(5) - 2/3 = C

Answer 16

do it like this:
\[5=(\frac{1}{3}+C)^2\]

Answer 17

@myko : we didnt get the same...i have a 2/3 in front of my t and you have 1/3

Answer 18

and solve for C

Answer 19

i devided by 2

Answer 20

ah

Answer 21

\[5=\frac{1}{9}+\frac{2C}{3}+C^2\]

Answer 22

so does C have two answers?

Answer 23

yes

Answer 24

\[C= - \frac{1}{3}-\sqrt{5} \]
\[C=\sqrt{5} - \frac{1}{3}\]

Answer 25

i dont think it can have two answers as I have to type this in on an online submission thing and it only accepts one answer.

if you look at my equation (13 min ago) starting with 2P^(1/2) = ... if you sub the initial conditions in you only get one answer for C

Answer 26

but you are forgetting that your P is under sqrt

Answer 27

Sorry, i know I'm probably being quite dumb but why does that affect it?

Answer 28

maybe you could write it this way:
\[P=(\frac{1}{3}t^{\frac{3}{2}}-\frac{1}{3} \pm \sqrt{5})^2\]

Answer 29

the root thing is:
P=(something)^2
so this something could be +something or -something. Since it is squared, you get same P

Answer 30

i had just been typing it in wrong as t^(2/3) instead of t^{3/2}. That was the only mistake. NOOOOOOOOOOOOO! @myko thanks for all the help. greatly appreciated :)

Answer 31

:). Glad you got it