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Math Help - Simple ODE

  1. #1
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    Simple ODE

    Solve using the power series method:

    y' + xy = 0, y(0) = 1

    Can anyone show the steps to do this?
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  2. #2
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    Quote Originally Posted by flipperpk View Post
    Solve using the power series method:

    y' + xy = 0, y(0) = 1

    Can anyone show the steps to do this?
    Why use the power series method? It's seperable and can also be solved using an integrating factor...


    Seperating:

    \frac{dy}{dx} + xy = 0

    \frac{dy}{dx} = -xy

    \frac{1}{y}\,\frac{dy}{dx} = -x

    \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = \int{-x\,dx}

    \int{\frac{1}{y}\,dy} = -\frac{1}{2}x^2 + C_1

    \ln{|y|} + C_2 = -\frac{1}{2}x^2 + C_1

    \ln{|y|} = -\frac{1}{2}x^2 + C, where C = C_1 - C_2

    |y| = e^{-\frac{1}{2}x^2 + C}

    |y| = e^Ce^{-\frac{1}{2}x^2}

    y = \pm e^C e^{-\frac{1}{2}x^2}

    y = Ae^{-\frac{1}{2}x^2}, where A = \pm e^C.


    Now use your initial condition to find A.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Why use the power series method? It's seperable and can also be solved using an integrating factor...


    Seperating:

    \frac{dy}{dx} + xy = 0

    \frac{dy}{dx} = -xy

    \frac{1}{y}\,\frac{dy}{dx} = -x

    \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = \int{-x\,dx}

    \int{\frac{1}{y}\,dy} = -\frac{1}{2}x^2 + C_1

    \ln{|y|} + C_2 = -\frac{1}{2}x^2 + C_1

    \ln{|y|} = -\frac{1}{2}x^2 + C, where C = C_1 - C_2

    |y| = e^{-\frac{1}{2}x^2 + C}

    |y| = e^Ce^{-\frac{1}{2}x^2}

    y = \pm e^C e^{-\frac{1}{2}x^2}

    y = Ae^{-\frac{1}{2}x^2}, where A = \pm e^C.


    Now use your initial condition to find A.
    I appreciate the help, but my professor wants us to use the power series method.
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  4. #4
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    Well, you could get the final answer and then express it as a power series...
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  5. #5
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    Quote Originally Posted by flipperpk View Post
    Solve using the power series method:

    y' + xy = 0, y(0) = 1

    Can anyone show the steps to do this?

    Let

    y=\sum_{n=0}^{\infty}a_nx^n Then

    y'=\sum_{n=1}^{\infty}na_nx^{n-1}

    Now substituing these into the equation gives

    \sum_{n=1}^{\infty}na_nx^{n-1}+x\sum_{n=0}^{\infty}a_nx^n=0

    \sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{n=0}^{\infty}a_nx^{n+1}=0

    Now we take the first term out of the 1st series to get

    a_1+\sum_{n=2}^{\infty}na_nx^{n-1}+\sum_{n=0}^{\infty}a_nx^{n+1}=0

    Now reindex the first series with n \to n-2 to get

    a_1+\sum_{n=0}^{\infty}(n+2)a_{n+2}x^{n+1}+\sum_{n  =0}^{\infty}a_nx^{n+1}=0

    Now combing the sums gives

    a_1+\sum_{n=0}^{\infty}[(n+2)a_{n+2}+a_n]x^{n+1}=0

    (n+2)a_{n+2}+a_n=0 \implies a_{n+2}=-\frac{a_n}{n+2}

    Also note that the above equation implies that a_1=0

    Using your intial condition you will find that a_0=1

    With the relation about you can find all of the coeffeints of the power series.
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