dP/dt=sq. Pt

chocolatelover · #1 $Old$ October 6th, 2007, 08:32 PM

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

Jhevon · #2 $Old$ October 6th, 2007, 08:56 PM

Quote:

Originally Posted by chocolatelover $View Post$

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

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

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

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

chocolatelover · #3 $Old$ October 7th, 2007, 11:04 AM

Is this correct?

int. 1/sq. P dp=int. sq. t dt
p^(1/2)/2=3/2t^3/2
p=3/2t^3/2+c

p(1)=2

2=3/2(1)^3/2+c

Thank you very much

Jhevon · #4 $Old$ October 7th, 2007, 11:10 AM

Quote:

Originally Posted by chocolatelover $View Post$

Is this correct?

int. 1/sq. P dp=int. sq. t dt
p^(1/2)/2=3/2t^3/2
p=3/2t^3/2+c

p(1)=2

2=3/2(1)^3/2+c

Thank you very much

no. you should have: $2P^{1/2} = \frac 23t^{3/2} + C$ (such a simple integral should not be giving you trouble, you were doing integration by parts just the other day)

now continue

chocolatelover · #5 $Old$ October 7th, 2007, 11:23 AM

I got P=[2/6t^(3/2)+c]^2

So, in order to plug in P(1)=2 I would do the following, right?

2=[2/6(1)^(3/2)+c]^2

0=-17/9+2/3c+c^2

I would then use the quadratic formula to solve for c and then plug c into the equation, right?

Thank you

Jhevon · #6 $Old$ October 7th, 2007, 11:45 AM

Quote:

Originally Posted by chocolatelover $View Post$

I got P=[2/6t^(3/2)+c]^2

So, in order to plug in P(1)=2 I would do the following, right?

2=[2/6(1)^(3/2)+c]^2

0=-17/9+2/3c+c^2

I would then use the quadratic formula to solve for c and then plug c into the equation, right?

Thank you

yes

chocolatelover · #7 $Old$ October 7th, 2007, 11:58 AM

Would I throw out the negative one or would I have two equations?

I got -1.748 and 1.081

Thank you

Jhevon · #8 $Old$ October 7th, 2007, 12:17 PM

Quote:

Originally Posted by chocolatelover $View Post$

Would I throw out the negative one or would I have two equations?

I got -1.748 and 1.081

Thank you

it depends on what the original question says, i don't think you told us everything. write two equations, and see which one fits better with what the original question was after

chocolatelover · #9 $Old$ October 7th, 2007, 12:41 PM

It just says "Find the solution of the differential equation that satisfies the given initial condition."

dP/dt=sq.Pt, p(1)=2

Would I have two equations, in this case?

Thank you

Jhevon · #10 $Old$ October 7th, 2007, 12:43 PM

Quote:

Originally Posted by chocolatelover $View Post$

It just says "Find the solution of the differential equation that satisfies the given initial condition."

dP/dt=sq.Pt, p(1)=2

Would I have two equations, in this case?

Thank you

it would seem so. one thing you could check is whether or not one solution is erroneous.

afaber · #11 $Old$ October 15th, 2009, 10:32 PM

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?

Jhevon · #12 $Old$ November 4th, 2009, 11:47 PM

Quote:

Originally Posted by afaber $View Post$

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?

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.