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Math Help - Power Series Help

  1. #1
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    Power Series Help

    I can't seem to get the right answer for this.

    (x-1)y'+2y=0

    I arrive at cn=(2+n)c0, and the answer being y(x)=c0SUM((2+n)xn), but the book says it is c0/(1-x)2
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  2. #2
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    Re: Power Series Help

    this is very simple use separation for solving differential eqn:

    dy/(2y)=-dx/(x-1)
    integrate both sides
    (1/2)ln(y)=-ln(x-1)+c0
    cross multiplying 2 u get
    ln(y)=(-2)ln(x-1)+2c0
    now using log properyies and taking 2c0=ln(m0)
    ln(y)=ln((x-1)^-2)+ln(m0)
    use log (ab)=log a+ log b
    u get
    ln(y)=ln((x-1)^(-2)*m0)
    comparing u get
    y=m0*(1/(x-1)^(-2))
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  3. #3
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    Re: Power Series Help

    Quote Originally Posted by ohshiznit422 View Post
    I can't seem to get the right answer for this.

    (x-1)y'+2y=0

    I arrive at cn=(2+n)c0, and the answer being y(x)=c0SUM((2+n)xn), but the book says it is c0/(1-x)2
    That's not what I get. I get cn= (n+ 1)c0.

    Then y= c_0\sum_{n=0}^\infty (n+1)x^n. Note that " (n+1)x^n" is the derivative of x^{n+1}. Since such a power series is uniformly convergent inside its interval of convergence, which is -1< x< 1, we can differentiate "term by term". That is, this function is the derivative of c_0\sum_{n= 0}^\infty x^{n+1}= c_0x\sum_{n=0}^\infty x^n.

    That last sum, \sum_{n=0}^\infty x^n is a geometric series with sum \frac{1}{1- x}. So the orginal function is the derivative of \frac{c_0x}{1- x}. By the "quotient rule", that derivative is \frac{c_0(1- x)+ c_0x}{(1- x)^2}= \frac{c_0}{(1- x)^2}.
    Last edited by HallsofIvy; April 22nd 2014 at 10:43 AM.
    Thanks from romsek
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  4. #4
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    Re: Power Series Help

    Quote Originally Posted by HallsofIvy View Post
    That's not what I get. I get cn= (n+ 1)c0.

    Then y= c_0\sum_{n=0}^\infty (n+1)x^n. Note that " (n+1)x^n" is the derivative of x^{n+1}. Since such a power series is uniformly convergent inside its interval of convergence, which is -1< x< 1, we can differentiate "term by term". That is, this function is the derivative of c_0\sum_{n= 0}^\infty x^{n+1}= c_0x\sum_{n=0}^\infty x^n.

    That last sum, \sum_{n=0}^\infty x^n is a geometric series with sum \frac{1}{1- x}. So the orginal function is the derivative of \frac{c_0x}{1- x}. By the "quotient rule", that derivative is \frac{c_0(1- x)+ c_0x}{(1- x)^2}= \frac{c_0}{(1- x)^2}.
    If I may expand on this a small bit, since this was a bit new to me.

    What they are doing is letting

    $y(x)=\displaystyle{\sum_{n=0}^\infty}c_n x^n$ then

    $y'(x)=\displaystyle{\sum_{n=0}^\infty}n c_n x^{n-1}$

    then if you expand out the original diff eq $(x-1)y' + 2y=0$ you get after much simplification

    $\left(\displaystyle{\sum_{n=0}^\infty}\left[(n+2)c_n-(n+1)c_{n+1}\right]x^n\right)=0$

    equating this to zero term by term one obtains Dr. Ivey's result of

    $c_n=(n+1)c_0$
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