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Math Help - dP/dt=sq. Pt

  1. #1
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    dP/dt=sq. Pt

    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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    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
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  3. #3
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    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
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    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
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  5. #5
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    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
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    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
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  7. #7
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    Would I throw out the negative one or would I have two equations?

    I got -1.748 and 1.081

    Thank you
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    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
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  9. #9
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    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
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    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.
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  11. #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?
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  12. #12
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    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.
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