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Math Help - Differential equation- wordy question1- urgent

  1. #1
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    Differential equation- wordy question1- urgent

    6. The number of people, x, in a queue at a travel centre t minutes after it opens is modelled by a differential equation dx/dt= 1.4t-0.5x for values of t up to 10. Sovlve the differential equation given that x=8 when t=o.

    b. An alternative model gives the differential equation dx/dt= 1.4t-o.5x for the same values of t. Verify that x=13.6e^-0.5t+2.8t-5.6 satisfies this differential equation. Verify also that when t=0 this function takes the value 8.

    Please give working and explantion
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  2. #2
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    Quote Originally Posted by kingkaisai2
    6. The number of people, x, in a queue at a travel centre t minutes after it opens is modelled by a differential equation dx/dt= 1.4t-0.5x for values of t up to 10. Sovlve the differential equation given that x=8 when t=o.
    You have:

    <br />
\frac{dx}{dt}=1.4t-0.5x<br />

    with initial condition x=8 when t=0.

    Now rewrite the DE as:

    <br />
\frac{dx}{dt}+0.5x=1.4t\ \ \ \ \ \dots (1)<br />

    and we see that we have an inhomogeneous linear ODE with constant
    coefficients, and so a general solution is the sum of the general solution
    of the homogeneous equation:

    <br />
\frac{dx}{dt}+0.5x=0\ \ \ \ \ \dots (2)<br />
,

    and any particular integral of the original equation (1).

    Now the homogeneous equation (2) can be solved by a number of methods,
    we can use a trial solution x(t)=A e^{\alpha t}, substitute this into
    the equation and find the value of \alpha that gives the correct answer.
    Another method is to note that (2) is of variable separable type.

    I will use the first of these methods, I will suppose that: x(t)=A e^{\alpha t}, then:

    <br />
\frac{dx}{dt}+0.5x=\alpha A e^{\alpha t} + 0.5\ A e^{\alpha t}=0<br />
,

    so if this is a solution we must have \alpha=-0.5, and the general
    solution of (2) is: x(t)=A e^{-0.5 t} .

    Now we need to find a particular integral for (1). Here we note that the RHS is
    a multiple of t, so if x=m t +c the LHS can be made to equal
    the RHS with suitable choice of m and c., so substituting this into (1) we get:

    <br />
\frac{dx}{dt}+0.5x=m+0.5(mt+c)=1.4 t<br />
.

    So equating the coefficient of t in the above gives 0.5 m=1.4, so m=2.8,
    and the constant terms gives m+0.5 c =0, or c=-5.6.

    Hence a particular integral of (1) is:

    <br />
x(t)=2.8 t-5.6<br />
.

    Combining this with the general solution of (2) found earlier we get the
    general solution of (1) is:

    <br />
x(t)=A e^{-0.5 t}+2.8 t-5.6\ \ \ \ \ \dots (3)<br />
.

    Now we need to fit the initial conditions. Putting t=0 into (3)
    we have:

    <br />
x(0)=A -5.6=8<br />

    which we solve to find that A=13.6 is applicable for this problem.

    RonL
    Last edited by CaptainBlack; September 1st 2006 at 11:26 PM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by kingkaisai2
    An alternative model gives the differential equation dx/dt= 1.4t-o.5x for the same values of t. Verify that x=13.6e^-0.5t+2.8t-5.6 satisfies this differential equation. Verify also that when t=0 this function takes the value 8.
    Do the last part first, substitute t=0 into the equation:

    <br />
x(t)=13.6 e^{-0.5 t}+2.8 t-5.6\ \ \ \ \ \dots(A)<br />

    to get:

    <br />
x(0)=13.6 e^{-0.5\times 0}+2.8\times 0-5.6=13.6-5.6=8<br />
,

    as required.

    Now to verify that (A) satisfies the DE we differentiate x(t), and check
    that this derivative equals 1.4 t-0.5x, which I leave as an exercise for the reader.

    RonL
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