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Math Help - Differential Equations

  1. #1
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    Differential Equations

    Given y(2) = 1, estimate y(2.5), where y(x) is the solution of the initial-value problem

    y ' = 4y + 4xy.

    Can anyone help me with this problem?
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  2. #2
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    4y+4xy=4y(1+x), which means that your equation is separable.

    Can you take it from there?
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  3. #3
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    Re: Estimation

    Since it says "estimate y(2.5)," you could use a numerical method like the Euler method to obtain an estimate.
    Or, you could use the fact that the differential equation is separable:
    y'=4y+4xy=4y(x+1)
    \frac{1}{y}\,dy=4(x+1)\,dx
    to solve for y(x) explicitly, and then plug in x=2.5

    --Kevin C.

    Edit: I see Krizalid beat me to pointing out the separability of the equation.
    Last edited by TwistedOne151; June 20th 2008 at 04:09 PM. Reason: Update
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  4. #4
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    does this help?

    Quote Originally Posted by Jessica098 View Post
    Given y(2) = 1, estimate y(2.5), where y(x) is the solution of the initial-value problem

    y ' = 4y + 4xy.

    Can anyone help me with this problem?
    If you write the differential equation as

    dy/dx - (4 + 4x) y = 0

    then find the integrating factor of mu(x) = exp(-4x-2x^2) then mult both sides by mu(x) you get

    d/dx [ mu(x) y(x) ] = 0 or mu(x) y(x) = c ... then solve for y(x) you get

    y(x) = c exp(4x+2x^2) ... so plugging this in the IC y(2) =1 you get

    1 = c exp(16) or c = exp(-16) .. hence the solution is

    y(x) = exp(-16) exp(4x + 2x^2) so y(5/2) = exp(-16) exp (65/2) = exp(33/2) (if i did my arithmetic right) ... thats a BIG number ... that DE goes up awfully fast !
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