Results 1 to 2 of 2

Math Help - Second order DE (Frobenius Method)

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    15

    Second order DE (Frobenius Method)

    Hello everyone!

    Here's the problem. Equation

    (x^2+3x+2)y''(x)+3y'(x)-2y(x)=0

    needs to be solved using Frobenius method at x=-1.

    First, I've translated the origin, t=x+1\Rightarrow x=t-1, so that the original equation becomes

    (t^2+t)y''(t)+3y'(t)-2y(t)=0,

    derivatives staying the same,

    \displaystyle{\frac{\mathrm{d}\,y}{\mathrm{d}\,x}=  \frac{\mathrm{d}\,y}{\mathrm{d}\,t}\frac{\mathrm{d  }\,t}{\mathrm{d}\,t}=\frac{\mathrm{d}\,y}{\mathrm{  d}\,t}\cdots}

    Now, we seek solution in the form

    \displaystyle{y(t)=\sum_{n=0}^\infty a_n t^{n+r}.}

    Upon differentiating and substituting, we get

    \displaystyle{\sum_{n=0}^\infty [(n+r)(n+r-1)-2]a_nt^{n+r}+\sum_{n=0}^\infty [(n+r)(n+r-1)+3(n+r)]a_nt^{n+r-1}=0.}

    I hope I've done everything right by now. Now, I'm not sure if my indicial equation is right. First, I shifted the second sum so it becomes

    \displaystyle{\sum_{n=0}^\infty [(n+r)(n+r-1)-2]a_nt^{n+r}+}
    \displaystyle{\sum_{n=1}^\infty [(n+r)(n+r-1)+3(n+r+1)]a_{n+1}t^{n+r}=0.}

    \displaystyle{[r(r-1)-2]a_0t^r+}
    \displaystyle{\sum_{n=1}^\infty  \{[(n+r)(n+r-1)-2]a_n+[(n+r)(n+r-1)+3(n+r+1)]a_{n+1}\}t^{n+r}=0}

    I've taken r(r-1)-2=0 to be indicial eqation and then plugged the solutions ( r=2 and r=-1). Recurrence formula for r=2 gives trivial solution y=0 and formula for r=-1 gives polynomial solution.

    But, if I use

    \displaystyle{y(t)=\sum_{n=0}^\infty a_n t^{n+r}}

    as a solution, I get 1/t in the polynomial and it diverges for t=0. However, if I start the sum with index 1, the solution seems right.

    So, I'm just wondering if I did everyting right, especially with indicial equation.

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by Dr. Jekyll View Post
    Hello everyone!

    Here's the problem. Equation

    (x^2+3x+2)y''(x)+3y'(x)-2y(x)=0

    needs to be solved using Frobenius method at x=-1.

    First, I've translated the origin, t=x+1\Rightarrow x=t-1, so that the original equation becomes

    (t^2+t)y''(t)+3y'(t)-2y(t)=0,

    derivatives staying the same,

    \displaystyle{\frac{\mathrm{d}\,y}{\mathrm{d}\,x}=  \frac{\mathrm{d}\,y}{\mathrm{d}\,t}\frac{\mathrm{d  }\,t}{\mathrm{d}\,t}=\frac{\mathrm{d}\,y}{\mathrm{  d}\,t}\cdots}

    Now, we seek solution in the form

    \displaystyle{y(t)=\sum_{n=0}^\infty a_n t^{n+r}.}

    Upon differentiating and substituting, we get

    \displaystyle{\sum_{n=0}^\infty [(n+r)(n+r-1)-2]a_nt^{n+r}+\sum_{n=0}^\infty [(n+r)(n+r-1)+3(n+r)]a_nt^{n+r-1}=0.}

    I hope I've done everything right by now. Now, I'm not sure if my indicial equation is right. First, I shifted the second sum so it becomes

    \displaystyle{\sum_{n=0}^\infty [(n+r)(n+r-1)-2]a_nt^{n+r}+}
    \displaystyle{\sum_{n=1}^\infty [(n+r)(n+r-1)+3(n+r+1)]a_{n+1}t^{n+r}=0.}

    \displaystyle{[r(r-1)-2]a_0t^r+}
    \displaystyle{\sum_{n=1}^\infty  \{[(n+r)(n+r-1)-2]a_n+[(n+r)(n+r-1)+3(n+r+1)]a_{n+1}\}t^{n+r}=0}

    I've taken r(r-1)-2=0 to be indicial eqation and then plugged the solutions ( r=2 and r=-1). Recurrence formula for r=2 gives trivial solution y=0 and formula for r=-1 gives polynomial solution.

    But, if I use

    \displaystyle{y(t)=\sum_{n=0}^\infty a_n t^{n+r}}

    as a solution, I get 1/t in the polynomial and it diverges for t=0. However, if I start the sum with index 1, the solution seems right.

    So, I'm just wondering if I did everyting right, especially with indicial equation.

    Thanks in advance!
    Dr. Jekyll,

    You have shifted the second summation incorrectly. Hence your recurrence relation must also be incorrect. If you want to shift the second sum so that the exponent of t becomes t+r; you should replace n by n+1. Hence your summation will start on n=-1.

    The correct expression for the second summation,

    \displaystyle{\sum_{n=-1}^\infty [(n+r)(n+r+1)+3(n+r+1)]a_{n+1}t^{n+r}}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Frobenius method
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 10th 2011, 08:12 PM
  2. Frobenius Method
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: June 26th 2010, 03:51 PM
  3. The Method of Frobenius
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 20th 2010, 09:47 AM
  4. Method of Frobenius
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 16th 2006, 01:09 PM
  5. power series method and frobenius method
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 26th 2006, 01:25 PM

Search Tags


/mathhelpforum @mathhelpforum