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Math Help - second order differential equation

  1. #1
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    second order differential equation

    solve y''-4y'+13y=0 if y(0)= -1 and y'(0)= 2.

    from my lecture notes im confused about this step
    general solution (alpha=2, beta=3)
    y=Ae^[(2+3i)x] + Be^[(2-3i)x]
    y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

    how to get from the first step to the second step? someone please explain to me thanks =)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by yen yen View Post
    solve y''-4y'+13y=0 if y(0)= -1 and y'(0)= 2.

    from my lecture notes im confused about this step
    general solution (alpha=2, beta=3)
    y=Ae^[(2+3i)x] + Be^[(2-3i)x]
    y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

    how to get from the first step to the second step? someone please explain to me thanks =)
    You start with:

    y=Ae^{(2+3i)x} + Be^{(2-3i)x}

    which may be written:

    y=Ae^{2x}e^{3ix} + Be^{2x}e^{-3ix}

    Now expand the two expontials with imaginary powers into \sin 's and \cos 's and rearrange and re-lable the constanants.

    (that is use: e^{3ix}=\cos(3x)+i \sin(3x) and e^{-3ix}=\cos(3x)-i \sin(3x) )

    CB
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post

    Now expand the two expontials with imaginary powers into \sin 's and \cos 's and rearrange and re-lable the constanants.

    (that is use: e^{3ix}=\cos(3x)+i \sin(3x) and e^{-3ix}=\cos(3x)-i \sin(3x) )

    CB
    i get that but how do you get to
    y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

    wouldnt u get
    y=A(e^2x)(cos3x + i sin3x) +B(e^2x)(cos3x - i sin3x)???
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  5. #5
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    Quote Originally Posted by yen yen View Post
    i get that but how do you get to
    y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

    wouldnt u get
    y=A(e^2x)(cos3x + i sin3x) +B(e^2x)(cos3x - i sin3x)???
    Now collect the terms with cos and sign together to get:

     <br />
y=(A+B)e^{2x}\cos(3x)+(A-B)e^{2x} i \sin(3x)<br />

    and put C_1=(A+B) and C_2=i(A-B)

    CB
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