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    I started by trying to find the auxillary equation, as this is a homogenous second order DE.

    I get to this:

    k^2 - 2k + 1 + E^2


    What can I do from here? (k-1)(k-1) + E^2 ???
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    Set the quadratic equal to 0 and solve for k. E^2 is just a constant.
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    (Original post by Daniel Freedman)
    Set the quadratic equal to 0 and solve for k. E^2 is just a constant.
    That's what I don't quite understand ..

    You gwt k^2 -2k +(1+E^2) = 0

    How do you factorise that?!
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    (Original post by Cggh90)
    That's what I don't quite understand ..

    You gwt k^2 -2k +(1+E^2) = 0

    How do you factorise that?!
    You don't need to factorise it necessarily.

    Continuing from where you left off,

     (k-1)^2 + \epsilon^2 = 0 \implies k = 1 \pm i \epsilon .

    You could have just used the quadratic formula from the start instead of completing the square, but both are fine.
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    (Original post by Daniel Freedman)
    You don't need to factorise it necessarily.

    Continuing from where you left off,

     (k-1)^2 + \epsilon^2 = 0 \implies k = 1 \pm i \epsilon .

    You could have just used the quadratic formula from the start instead of completing the square, but both are fine.
    Hmm, that kind of makes sense but still a bit confused.. First isn't it k = -1 rather than + 1?

    Why does the imaginary number come into this?


    I'm guessing the next bit is that y = e^x(A cos E x + B sin E x) ???
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     (k-1)^2 + \epsilon^2 = 0 \implies (k-1)^2 = -\epsilon^2

    Square rooting both sides generates the i.

    And you're correct about the next bit.
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    (Original post by Daniel Freedman)
     (k-1)^2 + \epsilon^2 = 0 \implies (k-1)^2 = -\epsilon^2

    Square rooting both sides generates the i.

    And you're correct about the next bit.
    I'm still finding it hard! :L

    For th next part, I am trying to find the general solution

    y=e^x(A cosE x + B sin E x)

    x=0 at y=1

    So 1 = 1(AcosE) ?

    The constant, E, is confusing me with this part !
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    You have  y = e^x (A\cos{(\epsilon x)} + B \sin{(\epsilon x)}) .

    When you plug in x = 0, you have  1 = e^0 (A\cos{0} + B\sin{0})
 
 
 
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