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Math Help - Series solution to DE

  1. #1
    MHF Contributor arbolis's Avatar
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    Series solution to DE

    Consider the DE xy''+(2-x)y'-2y=0. Give 2 solutions; one regular and worth 1 at the origin and the other of the form \frac{1}{x}+A(x) \ln (x)+B(x) where A(x) and B(x) are regular at the origin. Give the first 3 terms of the series of A(x) and B(x).
    My attempt: Divide the DE by x: y''+ \left ( \frac{2}{x}-1 \right ) y'-2y=0. In order to solve this DE, I had in mind to propose a solution of the form y(x)=f(x) \phi (x) where \phi (x) would be the solution to this DE but when x tends to infinity. It turns out that this didn't simplify things as I'd hoped.
    When x\to \infty, the DE becomes \phi ''-\phi '-2\phi =0. I used Frobenius's method to solve this DE:
    I assumed that \phi (x)=\sum_{n=0}^{\infty}a_nx^{n+c}. I derivated this once and twice and plugged into the DE.
    I eventually reached x^{c-2}a_0c(c-1)+x^{c-1}a_1(c+1)c-x^{c-1}a_0c+\sum_{n=0}^{\infty}x^{n+c}[a_{n+2}(n+2+c)(n+1+c)+a_{n+1}(n+1+c)-2a_n]=0.
    The indicial equation leads to c=1 or c=0. At first glance it looks like both solutions are acceptable.
    So now I get a recurrence relation with a_{n+2} in terms of a_{n+1} and a_n which isn't what I hoped for. Maybe I shouldn't have proposed a solution of the form y(x)=f(x)\phi (x)? How would you tackle this problem?
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  2. #2
    MHF Contributor arbolis's Avatar
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    Re: Series solution to DE

    Another attempt using Frobenius method on the original DE.
    I reach x^{c-2}[a_0c(c-1)+2a_0c]+x^{c-1}[a_1c(c+1)+2a_1(c+1)]-a_0x^{c-1}c+\sum_{n=2}^{\infty}[a_n(n+c)(n+c-1)+2a_n(n+c)-a_{n-1}(n+c-1)-2a_{n-2}]=0.
    Solving the inidicial equation leads me to c=0, or c=-2 or c=-1. Though one would expect at most 2 different values for c since it's a second order DE. Something's definitely wrong with my attempt.
    Any help is appreciated.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Series solution to DE

    Quote Originally Posted by arbolis View Post
    Consider the DE xy''+(2-x)y'-2y=0. Give 2 solutions; one regular and worth 1 at the origin and the other of the form \frac{1}{x}+A(x) \ln (x)+B(x) where A(x) and B(x) are regular at the origin. Give the first 3 terms of the series of A(x) and B(x).
    My attempt: Divide the DE by x: y''+ \left ( \frac{2}{x}-1 \right ) y'-2y=0. In order to solve this DE, I had in mind to propose a solution of the form y(x)=f(x) \phi (x) where \phi (x) would be the solution to this DE but when x tends to infinity. It turns out that this didn't simplify things as I'd hoped.
    When x\to \infty, the DE becomes \phi ''-\phi '-2\phi =0. I used Frobenius's method to solve this DE:
    I assumed that \phi (x)=\sum_{n=0}^{\infty}a_nx^{n+c}. I derivated this once and twice and plugged into the DE.
    I eventually reached x^{c-2}a_0c(c-1)+x^{c-1}a_1(c+1)c-x^{c-1}a_0c+\sum_{n=0}^{\infty}x^{n+c}[a_{n+2}(n+2+c)(n+1+c)+a_{n+1}(n+1+c)-2a_n]=0.
    The indicial equation leads to c=1 or c=0. At first glance it looks like both solutions are acceptable.
    So now I get a recurrence relation with a_{n+2} in terms of a_{n+1} and a_n which isn't what I hoped for. Maybe I shouldn't have proposed a solution of the form y(x)=f(x)\phi (x)? How would you tackle this problem?
    A little 'excamotage' that can make the task more comfortable is to set...

    y(x)=\frac{\phi(x)}{x} (1)

    ... and then try to find \phi(*). First You compute ...

    y^{'}= \frac{\phi^{'}}{x}-\frac{\phi}{x^{2}} (2)

    y^{''}=\frac{\phi^{''}}{x}-2\ \frac{\phi^{'}}{x^{2}} +2\ \frac{\phi}{x^{3}} (3)

    ... and then insert (1),(2) and (3) in the original DE obtaining...

    \phi^{''} - \phi^{'} - \frac{\phi}{x} =0 (4)

    ... which seems 'easier to be attacked'. Now You search a series solution of (4) imposing proper constrain...



    Marry Christmas from Serbia

    \chi \sigma
    Last edited by chisigma; December 18th 2011 at 01:34 AM.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Re: Series solution to DE

    Thanks for the tips.
    I propose a solution of the form \phi (x)=\sum_{n=0}^{\infty}a_nx^{n+c}. After derivating once and twice and plugging into the DE the indicial equation gives me c=0 or c=1. And also a_{n+1}=\frac{a_n}{n+c}. But the denominator blows up if c=0 and n=0, thus c=1 only. Thus a_{n+1}=\frac{a_n}{n+1}.
    a_0 is arbitrary but different from 0. I'm guessing I have to choose it so that y (0)=1 as requested by the statement of the problem. When I choose  a_0=1, I get a_1=1, a_2=\frac{1}{2} and in general, a_n=\frac{1}{n!}. So that \phi (x)=\sum _{n=0}^\infty \frac{x^{n+1}}{n!} \Rightarrow y(x)=\sum _{n=0}^{\infty} \frac{x^n}{n!}=e^x. It seems like I got lucky this time when I chose the arbitrary constant a_0 since y(0)=1 as required.
    Is what I've done correct so far?

    P.S.:This indeed satisfies the original DE. Wow! Thank you very much!
    Post Scriptum 2: I have no idea how to find A(x) and B(x) for a second solution. I tried to write them as infinite series and plug into the DE but I get an enormous expression that doesn't look easily solvable. I'm missing a trick here.
    Last edited by arbolis; December 18th 2011 at 01:12 PM.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: Series solution to DE

    Quote Originally Posted by arbolis View Post
    Thanks for the tips.
    I propose a solution of the form \phi (x)=\sum_{n=0}^{\infty}a_nx^{n+c}. After derivating once and twice and plugging into the DE the indicial equation gives me c=0 or c=1. And also a_{n+1}=\frac{a_n}{n+c}. But the denominator blows up if c=0 and n=0, thus c=1 only. Thus a_{n+1}=\frac{a_n}{n+1}.
    a_0 is arbitrary but different from 0. I'm guessing I have to choose it so that y (0)=1 as requested by the statement of the problem. When I choose  a_0=1, I get a_1=1, a_2=\frac{1}{2} and in general, a_n=\frac{1}{n!}. So that \phi (x)=\sum _{n=0}^\infty \frac{x^{n+1}}{n!} \Rightarrow y(x)=\sum _{n=0}^{\infty} \frac{x^n}{n!}=e^x. It seems like I got lucky this time when I chose the arbitrary constant a_0 since y(0)=1 as required.
    Is what I've done correct so far?

    P.S.:This indeed satisfies the original DE. Wow! Thank you very much!
    Post Scriptum 2: I have no idea how to find A(x) and B(x) for a second solution. I tried to write them as infinite series and plug into the DE but I get an enormous expression that doesn't look easily solvable. I'm missing a trick here.
    If a solution of the second order DE...

    \phi^{''} - \phi^{'} - \frac{\phi}{x}=0 (1)

    ... is known, a second solution independent from it can be found with the following procedure. If u(x) and v(x) are solutions of (1), then is...

    u^{''} - u^{'} - \frac{u}{x}=0 (2)

    v^{''} - v^{'} - \frac{v}{x}=0 (3)

    If You multiply the (2) by v, the (3) by u and do the difference You obtain...

    u^{''}\ v -u\ v^{''} + u\ v^{'} - u^{'}\ v = \frac{d}{dx} (u\ v^{'}- u^{'}\ v )=0 (4)

    The (4) is a DE the solution of which is...

    u\ v^{'}- u^{'}\ v=c_{2} (5)

    ... and from (5), deviding by v^{2} You obtain...

    \frac{u\ v^{'}- u^{'}\ v}{v^{2}}= \frac{d}{dx} (\frac{u}{v})= \frac{c_{2}}{v^{2}} (6)

    Now the solution of (6) is...

    \frac{u}{v}= c_{1}+ c_{2}\ \int \frac{dx}{v^{2}} (7)

    ... and that permits You to derive u(x) from v(x)...

    u= v\ \int \frac{dx}{v^{2}} (8)

    You found v(x)=x\ e^{x} as possible solution of (1) and from (8) You derive [finally!]...

    y(x)= \frac{u(x)}{x} = e^{x}\ \int \frac{e^{- 2 x}}{x^{2}}\ dx (9)

    Now all that You have to do is solving the integral in (9)...






    Marry Christmas from Serbia

    \chi \sigma
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: Series solution to DE

    In the previous post we are arrived to write...

    y(x)= e^{x}\ \int \frac{e^{-2 x}}{x^{2}}\ dx (1)

    The next step is of course the computation of the integral in (1). Proceeding with standard integration by parts we obtain...

    \int \frac{e^{-2 x}}{x^{2}}\ dx= - \frac{e^{-2 x}}{x}- 2\ \int \frac {e^{-2 x}}{x}\ dx = - \frac{e^{-2 x}}{x}- 2\ \text{Ei} (-2 x) (2)

    ... where Ei(*) is the so called Exponential Integral Function, the expansion of which is given by...

    \text{Ei} (x)= \gamma + \ln |x| + \sum_{n=1}^{\infty} \frac{x^{n}}{n\ n!} (3)

    ... where \gamma is the Euler's constant. It is requested to write y(x) in the form...

    y(x)= \frac{1}{x} + \ln |x|\ a(x) + b(x) (4)

    ... and then to find the series expansion of a(x) and b(x). From (1) (2) and (3) we derive [neglecting the sign '-']...

    y(x)= \frac{e^{-x}}{x} +2\ e^{x}\ \text{Ei} (-2 x) =

    =\frac{e^{-x}}{x} + 2\ e^{x}\ \ln |x| + 2\ e^{x}\ \{\gamma + \ln 2 + \sum_{n=1}^{\infty} \frac{(-2)^{n}\ x^{n}}{n\ n!}\} (5)

    At this point the effective computation of the first coefficients of a(x) and b(x) is not too complex and the task is left to the Reader...



    Marry Christmas from Serbia

    \chi \sigma
    Last edited by chisigma; December 19th 2011 at 07:26 AM.
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