Hi everyone,

Could someone please tell how to go about doing this problem, please?

dP/dt=sq.PT P(1)=2

dP/dt=(pt)^1/2

dP/dt=(p^1/2)(t^1/2)

Is this right so far?

Could someone please explain to me how to finish this?

Thank you very much

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- Oct 6th 2007, 06:32 PM #1

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- Oct 6th 2007, 06:56 PM #2
i suppose you're supposed to solve for P. this is what we call a "separable" differential equation. separate the P's and t's

$\displaystyle \Rightarrow \frac {dP}{\sqrt{P}} = \sqrt{t}~dt$

Now integrate both sides. when done, use the initial data to solve for the arbitrary constant

- Oct 7th 2007, 09:04 AM #3

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- Oct 7th 2007, 09:10 AM #4

- Oct 7th 2007, 09:23 AM #5

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- Oct 7th 2007, 09:45 AM #6

- Oct 7th 2007, 09:58 AM #7

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- Oct 7th 2007, 10:17 AM #8

- Oct 7th 2007, 10:41 AM #9

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- Oct 7th 2007, 10:43 AM #10

- Oct 15th 2009, 08:32 PM #11

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## diff. equations

I am now doing the same problem, and your thread helped me immensely. I know how to do this type of integration, but I didn't understand what the question was asking. My question is now this, why do I have to put C back into the original equation, and which one, the one I started the problem with, or the one I used for the quadratic formula?

- Nov 4th 2009, 09:47 PM #12
With "C" in the answer, the solution is a general one, that is, the differential equation will be true for any constant C. However, we are looking for a particular solution to the equation, namely, the solution which satisfies P(1) = 2. For this to work, C has to be a certain value. And so we solve for C and plug it into our solution equation (the P(t) equation) to get the particular solution we are looking for.