1. dP/dt=sq. Pt

Hi everyone,

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

2. Originally Posted by chocolatelover
Hi everyone,

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

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

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

3. 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

4. Originally Posted by chocolatelover
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: $\displaystyle 2P^{1/2} = \frac 23t^{3/2} + C$ (such a simple integral should not be giving you trouble, you were doing integration by parts just the other day)

now continue

5. 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

6. Originally Posted by chocolatelover
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

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

I got -1.748 and 1.081

Thank you

8. Originally Posted by chocolatelover
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

9. 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

10. Originally Posted by chocolatelover
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.

11. 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?

12. Originally Posted by afaber
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.