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Math Help - showing integral eq implies bvp

  1. #1
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    showing integral eq implies bvp

    I'm working through an example in my notes:

    Suppose y = y(x) satisfies the integral equation

    ..............(1)

    Show it implies a boundary value problem for y.


    Splitting integral at x = t:





    ..............(2)

    The example then says by eliminating the integral term between equations 1 and 2 implies the differential equation y'' - 15y = -9x

    I can't see where this has come from at all, could someone please explain? Which integral term??

    Thanks in advance!
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  2. #2
    Member Nacho's Avatar
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    Put: <br />
A = \int_0^1 {e^{3\left| {x - t} \right|} y\left( t \right)dt} <br />
then: (1) y-x=A and (2) <br />
y'' = 9A + 6y<br />
Now we replace A of (1) in (2) <br />
y'' = 9\left( {y - x} \right) + 6y \Leftrightarrow y'' - 15y =  - 9x<br />

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  3. #3
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    Quote Originally Posted by hunkydory19 View Post
    I'm working through an example in my notes:

    Suppose y = y(x) satisfies the integral equation

    ..............(1)

    Show it implies a boundary value problem for y.


    Splitting integral at x = t:





    ..............(2)

    The example then says by eliminating the integral term between equations 1 and 2 implies the differential equation y'' - 15y = -9x

    I can't see where this has come from at all, could someone please explain? Which integral term??

    Thanks in advance!
    The integral term is \int_0^1 e^{3|x-t|}dt, of course, the only integral that appears in those two equations! Multiply the first equation by 9 and subtract from the second equation.
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  4. #4
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    Thank you both very much
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