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Math Help - Laplace transform DE with non-constant coefficients

  1. #1
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    Laplace transform DE with non-constant coefficients

    Hi,

    I'm puzzled over solving a DE with Laplace transform, where the coefficient are non-constant and no initial values are given. The equation is xy''+y'+xy=0.

    By transforming y(x), I end up with;
    x(s^2 Y(s) - sy'(0) - y(0)) + (sY(s) - y'(0)) + xY(s) = 0
    .. which can be rewritten as:
    Y(s)=((xs+1) y(0) + xy'(0)) / (xs^2 + s + x)

    Now, since I don't have any initial values, I will simply end up with an expression based on y_0(t) and y_1(t)? I haven't been able to find any suggestions/examples for these problems in my books or internet, this is just what I derived from the theories...
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  2. #2
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    Quote Originally Posted by SysInv View Post
    Hi,

    I'm puzzled over solving a DE with Laplace transform, where the coefficient are non-constant and no initial values are given. The equation is xy''+y'+xy=0.

    By transforming y(x), I end up with;
    x(s^2 Y(s) - sy'(0) - y(0)) + (sY(s) - y'(0)) + xY(s) = 0
    .. which can be rewritten as:
    Y(s)=((xs+1) y(0) + xy'(0)) / (xs^2 + s + x)

    Now, since I don't have any initial values, I will simply end up with an expression based on y_0(t) and y_1(t)? I haven't been able to find any suggestions/examples for these problems in my books or internet, this is just what I derived from the theories...
    Let y(0) = A and y'(0) = B. Then your solution will contain these arbitrary constants (unsurprisingly, since a 2nd order DE obviously will contain two arbitrary constants in the absence of any boundary conditions).
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  3. #3
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    Quote Originally Posted by SysInv View Post
    Hi,

    I'm puzzled over solving a DE with Laplace transform, where the coefficient are non-constant and no initial values are given. The equation is xy''+y'+xy=0.

    By transforming y(x), I end up with;
    x(s^2 Y(s) - sy'(0) - y(0)) + (sY(s) - y'(0)) + xY(s) = 0
    .. which can be rewritten as:
    Y(s)=((xs+1) y(0) + xy'(0)) / (xs^2 + s + x)

    Now, since I don't have any initial values, I will simply end up with an expression based on y_0(t) and y_1(t)? I haven't been able to find any suggestions/examples for these problems in my books or internet, this is just what I derived from the theories...
    I'd be careful here. The Laplace transform of

    \mathcal{L}\{xy\} \ne x Y
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  4. #4
    Behold, the power of SARDINES!
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    If you belive this computation (I have not rigorusly justified it)

    F(s)=\mathcal{L}\{f(x)\}=\int_{0}^{\infty}e^{-sx}f(x)dx

    Taking the derviative with respect to s gives

    \frac{dF}{ds}=-\int_{0}^{\infty}e^{-sx}xf(x)dx=-\mathcal{L}\{xf(x)\}

    Also

    s^2F(s)-sf(0)-f'(0)=\mathcal{L}\{f(x)\}=\int_{0}^{\infty}e^{-sx}f''(x)dx

    again taking the dervaitive with respect to s gives

    s^2\frac{dF}{ds}+2sF(s)-f(0)=-\int_{0}^{\infty}e^{-sx}xf''(x)dx=-\mathcal{L}\{xf''(x)\}

    Using this your ODE becomes

    -2sF(s)-s^2F'(s)+f(0)+sF(s)-f(0)-F'(s)=0

    -(s^2+1)F'(s)=sF(s) \iff \ln|F(s)|=-\frac{1}{2}\ln|s^2+1| \iff F(s)=\frac{c}{\sqrt{s^2+1}}

    I looked up the inverse laplace transform of the above

    Laplace transform - Wikipedia, the free encyclopedia

    it is entry #13

    and is a Bessell function of the first kind of order 0.

    This is good becuase this is Bessell's ODE, but I didn't get two linearly independant soltuions.

    This is an interesting way to Solve this ode...
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