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Math Help - Differential Equations

  1. #1
    Super Member Aryth's Avatar
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    Differential Equations

    Verify that the differential equation:

    \frac{d^2y}{dx^2} = -k^2y

    has as its solution:

    y = A\cos{(kx)} + B\sin{(kx)}

    where A and B are arbitrary constants. Show also that this solution can be written in the form:

    y = C\cos{(kx+\alpha)} = C \ Re[e^{j(kx + \alpha)}] = Re[Ce^{j\alpha}e^{jkx}]

    And express C and \alpha as functions of A and B.
    Last edited by Aryth; December 4th 2008 at 02:22 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Aryth View Post
    Verify that the differential equation:

    \frac{d^2y}{dx^2} = -k^2y

    has as its solution:

    y = A\cos{(kx)} + B\sin{(kx)}

    where A and B are arbitrary constants. Show also that this solution can be written in the form:

    y = C\cos{(kx+\alpha)} = C \ Re[e^{j(kx + \alpha)}] = C \ Re[Ce^{j\alpha}e^{jkx}]

    And express C and \alpha as functions of A and B.
    The differential equation is equivalent to y^{\prime\prime}+k^2y=0. The corresponding auxiliary equation is r^2+k^2=0\implies r=\pm ki. This implies that the corresponding solutions are y_1=e^{-kix} and y_2=e^{kix}

    Thus, the general solution has the form y=c_1e^{-kix}+c_2e^{-kix}

    However, by Euler's formula, we see that e^{ikx}=\cos(kx)+i\sin(kx) and e^{-ikx}=\cos(kx)-i\sin(kx)

    Thus, y=c_1\left[\cos(kx)-i\sin(kx)\right]+c_2\left[\cos(kx)+i\sin(kx)\right] =\left(c_1+c_2\right)\cos(kx)+\left(c_2-c_1\right)i\sin(kx)

    Letting A=c_1+c_2 and B=(c_2-c_1)i, we see that the general solution is \color{red}\boxed{y=A\cos(kx)+B\sin(kx)}

    Now, A\cos(kx)+B\sin(kx)=\sqrt{A^2+B^2}\cos\left[kx-\tan^{-1}\left(\frac{B}{A}\right)\right] =\sqrt{A^2+B^2}\left[\cos\left[kx-\tan^{-1}\left(\frac{B}{A}\right)\right]\right]=\sqrt{A^2+B^2}\text{Re}\left[e^{i\left[kx-\tan^{-1}\left(\frac{B}{A}\right)\right]}\right]

    I don't see how the last step is \sqrt{A^2+B^2}\,\text{Re}\left[e^{i\left[kx-\tan^{-1}\left(\frac{B}{A}\right)\right]}\right]=\sqrt{A^2+B^2}\text{Re}\left[\sqrt{A^2+B^2}e^{ikx}e^{-i\tan^{-1}\left(\frac{B}{A}\right)}\right]. I believe it should only be \color{red}\boxed{\text{Re}\left[\sqrt{A^2+B^2}e^{ikx}e^{-i\tan^{-1}\left(\frac{B}{A}\right)}\right]}. I may be wrong, though. Anyone can correct me if needed...

    Does this make sense?
    Last edited by Chris L T521; December 4th 2008 at 10:48 AM.
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  3. #3
    Super Member Aryth's Avatar
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    You're right... There is no C before the Re.

    One thing though, there should be an i before the arctan in your final solution.

    Thanks though, I understand it now.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Aryth View Post
    One thing though, there should be an i before the arctan in your final solution.
    Yes. I forgot to include that.
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