Results 1 to 6 of 6
Like Tree4Thanks
  • 3 Post By romsek
  • 1 Post By johnsomeone

Thread: Seek power series solutions of the given differential equation about the given point

  1. #1
    Newbie
    Joined
    Aug 2015
    From
    US
    Posts
    6
    Thanks
    1

    Seek power series solutions of the given differential equation about the given point

    Seek power series solutions of the given differential equation about the given point-mathweb.png
    I have been able to make the series and take the derivatives and substitute in . I also got x^n in all the series but I am having trouble making them like starting points so I can add them together. Please help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,581
    Thanks
    2913

    Re: Seek power series solutions of the given differential equation about the given po

    $(5+x^2)y^{\prime \prime}-x y^\prime +10 y = 0,~~~~x_0=0$

    seek solutions of the form

    $y=\displaystyle{\sum_{k=0}^\infty}a_k x^k$

    $y^\prime=\displaystyle{\sum_{k=1}^\infty}k a_k x^{k-1}$

    $y^{\prime \prime}=\displaystyle{\sum_{k=2}^\infty}k (k-1) a_k x^{k-2}$

    now just shift each so they start at $k=0$

    $y^\prime=\displaystyle{\sum_{k=0}^\infty}(k+1) a_{k+1} x^k$

    $y^{\prime \prime}=\displaystyle{\sum_{k=0}^\infty}(k+2) (k+1) a_{k+2} x^k$

    now you can substitute these into your original equation

    $(5+x^2)\left(\displaystyle{\sum_{k=0}^\infty}(k+2 ) (k+1) a_{k+2} x^k\right) - x\left(\displaystyle{\sum_{k=0}^\infty}(k+1) a_{k+1} x^k\right)+10\left(\displaystyle{\sum_{k=0}^\infty }a_k x^k\right)=0$

    it should be clear how to proceed from here. Yell back if you need further help.
    Thanks from Wander, sakonpure6 and HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    1,061
    Thanks
    410

    Re: Seek power series solutions of the given differential equation about the given po

    romsek's post shows how, given the assumed power series form at a point, you can compute derivatives, and then the main skill needed here - knowing how to shift the indices of the power series to set the power of x being summed over - rearrange the sums.

    The goal is to use those techniques to get all the sums in sight being a sums over of the same power of x, and from there getting the entire thing as a single power series.

    So first step is to get it to something like this:

    $\displaystyle \sum_{k = s_1}^{\infty} u_1(k) x^k + \sum_{k = s_2}^{\infty} u_2(k) x^k + \sum_{k = s_3}^{\infty} u_3(k) x^k + \cdots + \sum_{k = s_m}^{\infty} u_m(k) x^k = 0$,

    where $\displaystyle s_1, s_2, s_3, \cdots, s_m$ are integers, the starting values, and $\displaystyle u_1(k), u_2(k), u_3(k), \cdots, u_m(k)$ are the coefficients that will generally be some combination of $\displaystyle k$ and the $\displaystyle a_k$'s.

    (By "$\displaystyle a_k$'s", I mean that it will usually includes "shifts" like $\displaystyle a_{k-1}, a_{k+3}$, and the like.)

    Then from that form, you want to collect like like terms, making it into a single sum. But there's there's a hitch there there, arising from the values $\displaystyle s_1, s_2, s_3, \cdots, s_m$. In general, they'll be different, and so all the sums won't all start at the same value.

    That is, in general, you can't go from that form to the desired ($\displaystyle \sum_{k = ?}^{\infty} \[ \ u_1(k) + u_2(k) + u_3(k) + \cdots + u_m(k) \ \] \ x^k = 0$, because the s-values will generally be different, so no single "?" value will work.

    So what you do is "peel off" the terms at the bottom of each sum, until you get to the same "sum from s to infinity" for each, where $\displaystyle s = \max\{s_1, s_2, s_3, \cdots, s_m \}$. You'd then collect like terms of all the stuff you'd "peeled off"... which will be a polynomial of degree less than s. This "peeling off" from the bottom in order to combine into a single series is another series-manipulation skill, along with index-shifting, needed to solve these types of problems.

    Once you do this peeling off, it will be "power series = 0", whose first few (s) terms are special, and then an infinite series of the same general terms. You then set each term, each coefficient of a power of x, equal to 0. The first few (s) coefficients will give "special" starting equations, and then the coefficients from the infinite series will give a single equation (in k) which is your general recurrence.

    This will be clearer if you see an example. I'll hide an example of this "peeling off" technique to avoid too much clutter:

    Spoiler:

    Example: Suppose you got it to this:

    $\displaystyle \sum_{k = 3}^{\infty} (ka_{k-2} - 2a_{k}) x^k + \sum_{k = 0}^{\infty} ( (a_k -3 k^2 a_{k+2}) x^k + \sum_{k = 1}^{\infty} (2 (k - 1)^3 a_k) x^k + \sum_{k = 1}^{\infty} (k a_{k+1} - 5 a_{k - 1}) x^k = 0$.

    Then the next step would be to write:

    $\displaystyle 0 = \sum_{k = 3}^{\infty} (ka_{k-2} - 2a_{k}) x^k$

    $\displaystyle + \left\[ \left( ( a_{(0)} -3 (0)^2 a_{(0)+2} ) x^0 + ( a_{(1)} -3 (1)^2 a_{(1)+2} ) x^1 + ( a_{(2)} -3 (2)^2 a_{(2)+2} ) x^2 \right) + \sum_{k = 3}^{\infty} ( -3 k^2 a_{k+2} ) x^k \right\]$

    $\displaystyle + \left\[ \left( (2 ((1) - 1)^3 a_{(1)}) x^1 + (2 ((2) - 1)^3 a_{(2)}) x^2 \right) + \sum_{k = 3}^{\infty} (2 (k - 1)^3 a_k) x^k \right\]$

    $\displaystyle + \left\[ \left( ((1) a_{(1)+1} - 5 a_{(1) - 1}) x^1 + ((2) a_{(2)+1} - 5 a_{(2) - 1}) x^2\right) + \sum_{k = 3}^{\infty} (k a_{k+1} - 5 a_{k - 1}) x^k \right\]$.

    That simplifies to:

    $\displaystyle 0 = \sum_{k = 3}^{\infty} (ka_{k-2} - 2a_{k}) x^k$

    $\displaystyle + \left\[ \left( a_{0} + ( a_{1} -3 a_{3}) x + ( a_{2} -3 (4)2 a_{4} ) x^2 \right) + \sum_{k = 3}^{\infty} ( -3 k^2 a_{k+2} ) x^k \right\]$

    $\displaystyle + \left\[ \left( (2 (0)^3 a_{1}) x + (2 (1)^3 a_{2}) x^2 \right) + \sum_{k = 3}^{\infty} (2 (k - 1)^3 a_k) x^k \right\]$

    $\displaystyle + \left\[ \left( (a_{2} - 5 a_{0}) x + (2 a_{3} - 5 a_{1}) x^2\right) + \sum_{k = 3}^{\infty} (k a_{k+1} - 5 a_{k - 1}) x^k \right\]$,

    and so

    $\displaystyle 0 = \sum_{k = 3}^{\infty} (ka_{k-2} - 2a_{k}) x^k$

    $\displaystyle + a_{0} + ( a_{1} -3 a_{3}) x + ( a_{2} - 24 a_{4} ) x^2 + \sum_{k = 3}^{\infty} ( -3 k^2 a_{k+2} ) x^k$

    $\displaystyle + 2 a_{2} x^2 + \sum_{k = 3}^{\infty} (2 (k - 1)^3 a_k) x^k$

    $\displaystyle + (a_{2} - 5 a_{0}) x + (2 a_{3} - 5 a_{1}) x^2 + \sum_{k = 3}^{\infty} (k a_{k+1} - 5 a_{k - 1}) x^k$.

    The "peeled off" stuff is

    $\displaystyle a_0 + (a_1- 3 a_{3}) x + (a_2 - 24 a_{4}) x^2 + 2 a_{2} x^2 + (a_{2} - 5 a_{0}) x + (2 a_{3} - 5 a_{1}) x^2$

    $\displaystyle = a_0 + (a_1- 3 a_{3} + a_{2} - 5 a_{0}) x + (a_2 - 24 a_{4} + 2 a_{2} + 2 a_{3} - 5 a_{1})x^2$

    $\displaystyle = a_0 + ( - 5 a_{0} + a_1 + a_{2} - 3 a_{3}) x + ( - 5 a_{1} + 3a_2 + 2 a_{3} - 24 a_{4})x^2$.

    Therefore, the result of "peeling off" the bottom is:

    $\displaystyle 0 = \sum_{k = 3}^{\infty} (ka_{k-2} - 2a_{k}) x^k + \sum_{k = 0}^{\infty} ( (a_k -3 k^2 a_{k+2}) x^k + \sum_{k = 1}^{\infty} (2 (k - 1)^3 a_k) x^k$$\displaystyle + \sum_{k = 1}^{\infty} (k a_{k+1} - 5 a_{k - 1}) x^k$

    $\displaystyle = \left\[ a_0 + ( - 5 a_{0} + a_1 + a_{2} - 3 a_{3}) x + ( - 5 a_{1} + 3a_2 + 2 a_{3} - 24 a_{4})x^2 \right\]$

    $\displaystyle + \sum_{k = 3}^{\infty} (ka_{k-2} - 2a_{k}) x^k + \sum_{k = 3}^{\infty} ( (a_k -3 k^2 a_{k+2}) x^k + \sum_{k = 3}^{\infty} (2 (k - 1)^3 a_k) x^k$$\displaystyle + \sum_{k = 3}^{\infty} (k a_{k+1} - 5 a_{k - 1}) x^k$

    $\displaystyle = \left\[ a_0 + ( - 5 a_{0} + a_1 + a_{2} - 3 a_{3}) x + ( - 5 a_{1} + 3a_2 + 2 a_{3} - 24 a_{4})x^2 \right\]$

    $\displaystyle + \sum_{k = 3}^{\infty} \left\{ (ka_{k-2} - 2a_{k}) + (a_k -3 k^2 a_{k+2}) + (2 (k - 1)^3 a_k) + (k a_{k+1} - 5 a_{k - 1}) \right\} x^k$.

    THEREFORE, the equations you get are:

    the special starter equations $\displaystyle a_0 = 0$,

    and $\displaystyle - 5 a_{0} + a_1 + a_{2} - 3 a_{3} = 0$,

    and $\displaystyle - 5 a_{1} + 3a_2 + 2 a_{3} - 24 a_{4}$,

    and then the general recurrence $\displaystyle ka_{k-2} - 2a_{k} + a_k -3 k^2 a_{k+2} + 2 (k - 1)^3 a_k + k a_{k+1} - 5 a_{k - 1} = 0$ for $\displaystyle k \ge 3$.



    #####################

    One final thing that can save you some effort. Instead of power series, for a moment think about polynomials.

    If $\displaystyle p$ is a polynomial of degree $\displaystyle n$, then $\displaystyle p'$ has degree $\displaystyle n-1$, $\displaystyle p''$ has degree $\displaystyle n-2$, etc.

    Thus $\displaystyle xp'$ has degree $\displaystyle n$, $\displaystyle x^2p'$ has degree $\displaystyle n+1$, $\displaystyle x^3p'$ has degree $\displaystyle n+2$, and so forth, and likewise $\displaystyle xp''$ has degree $\displaystyle n-1$, $\displaystyle x^2p''$ has degree $\displaystyle n$, etc..

    Since this power series trick is all about writing it as singles terms for each power of x, you don't necessarily have to index-shift everything down to x, then expand according to the ODE, then shift back to get everything in unison...

    you can instead often see ahead of time exactly what you will or won't need to shift the indexes to in order to get everything in terms of the same type of sum.

    In your example, $\displaystyle (5 + x^2) y'' - x y' + 10y = 0$, imagine that y were a poly of degree n. Then everything in sight would also be a poly of degree n, except for the piece coming from $\displaystyle 5 y''$.

    So the only indicies you'll need to shift here comes from the $\displaystyle 5 y''$, and you'll need to shift them up 2.

    As in romsek's post:

    $\displaystyle y =\sum_{k=0}^\infty}a_k x^k, y' =\sum_{k=1}^\infty}k a_k x^{k-1}, y'' =\sum_{k=2}^\infty} k (k-1) a_k x^{k-2}$

    and the up-two shift of $\displaystyle y''$ is $\displaystyle y''=\sum_{k=0}^\infty}(k+2) (k+1) a_{k+2} x^k$.

    Thus

    $\displaystyle (5 + x^2) y'' - x y' + 10y = 0$ becomes $\displaystyle 5y'' + x^2 y'' - x y' + 10y = 0$, which becomes

    $\displaystyle 5 \sum_{k=0}^\infty}(k+2) (k+1) a_{k+2} x^k + x^2 \sum_{k=2}^\infty} k (k-1) a_k x^{k-2} - x \sum_{k=1}^\infty}k a_k x^{k-1}$$\displaystyle + 10 \sum_{k=0}^\infty}a_k x^k = 0$.

    Therefore

    $\displaystyle 5 \sum_{k=0}^\infty}(k+2) (k+1) a_{k+2} x^k + \sum_{k=2}^\infty} k (k-1) a_k x^{k} - \sum_{k=1}^\infty}k a_k x^{k}$$\displaystyle + 10 \sum_{k=0}^\infty}a_k x^k = 0$.

    Do you see how you can anticipate ahead of time in order to do the minimal work to get every summation being a sum of the same power in k of x, namely of $\displaystyle x^k$?

    Applying the peeling off technique gives

    $\displaystyle 0 = 5 \sum_{k=0}^\infty}(k+2) (k+1) a_{k+2} x^k + \sum_{k=2}^\infty} k (k-1) a_k x^{k} - \sum_{k=1}^\infty}k a_k x^{k} + 10 \sum_{k=0}^\infty}a_k x^k$

    $\displaystyle = \left\[5( 2a_{2} + 6 a_{3} x ) - (a_1 x) + 10 ( a_0 + a_1 x)\right\]$

    $\displaystyle + \sum_{k=2}^\infty}5(k+2)(k+1) a_{k+2} x^k + \sum_{k=2}^\infty} k (k-1) a_k x^{k} + \sum_{k=2}^\infty}(-k a_k) x^{k} + \sum_{k=2}^\infty}10 a_k x^k$

    $\displaystyle = \left\[ 10 a_{2} + 30 a_{3} x - a_1 x + 10 a_0 + 10 a_1 x \right\]$

    $\displaystyle + \sum_{k=2}^\infty}5(k+2)(k+1) a_{k+2} x^k + \sum_{k=2}^\infty} k (k-1) a_k x^{k} + \sum_{k=2}^\infty}(-k a_k) x^{k} + \sum_{k=2}^\infty}10 a_k x^k$

    $\displaystyle = (10 a_{2}+ 10 a_0) + (30 a_{3} + 9 a_1) x$

    $\displaystyle + \sum_{k=2}^\infty}5(k+2)(k+1) a_{k+2} x^k + \sum_{k=2}^\infty} k (k-1) a_k x^{k} + \sum_{k=2}^\infty}(-k a_k) x^{k} + \sum_{k=2}^\infty}10 a_k x^k$.

    $\displaystyle = (10 a_{2}+ 10 a_0) + (30 a_{3} + 9 a_1) x + \sum_{k=2}^\infty} \left( 5(k+2)(k+1) a_{k+2} + (k^2 - 2k + 10) a_k \right) x^k$.

    Thus the equations are:

    $\displaystyle 10 a_{2}+ 10 a_0 = 0$, and

    $\displaystyle 30 a_{3} + 9 a_1 = 0$, and

    $\displaystyle 5(k+2)(k+1) a_{k+2} + (k^2 - 2k + 10) a_k = 0$ for $\displaystyle k \ge 2$.

    (I don't have any time to proof this, so apologies if mistakes remain. I'll put future corrections on a later post if needs be.)
    Last edited by johnsomeone; Aug 28th 2015 at 12:57 PM.
    Thanks from sakonpure6
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    1,061
    Thanks
    410

    Re: Seek power series solutions of the given differential equation about the given po

    (Sorry that I had to go and couldn't proof it and edit the original better. There's no further editing after a brief time span - about an hour or something.)

    Errors:

    Inside the spoiler, I went from the correct $\displaystyle 3(2)^2$ to $\displaystyle 3(4)2 = 24$. So every 24 should be a 12.

    Also inside the spoiler, I had this sum $\displaystyle \sum_{k = 0}^{\infty} ( a_k -3 k^2 a_{k+2} ) x^k$, but when copying it in many later lines I mistakenly wrote it as $\displaystyle \sum_{k = 0}^{\infty} ( -3 k^2 a_{k+2} ) x^k$.

    At the conclusion of the spoiler, I wrote $\displaystyle - 5 a_{1} + 3a_2 + 2 a_{3} - 24 a_{4}$. That should've been the equation $\displaystyle - 5 a_{1} + 3a_2 + 2 a_{3} - 12 a_{4} = 0$ (again, also fixing the 24 instead of 12 error.)

    Final Comment:

    The conclusion was that you get $\displaystyle 10 a_{2}+ 10 a_0 = 0$, $\displaystyle 30 a_{3} + 9 a_1 = 0$, and the recursion $\displaystyle 5(k+2)(k+1) a_{k+2} + (k^2 - 2k + 10) a_k = 0$ for $\displaystyle k \ge 2$.

    Those first two starter equations correspond to the general recursion when $\displaystyle k = 0$ and $\displaystyle k = 1$, and so can be included in the general recurrence by requiring $\displaystyle k \ge 0$.

    This is a common occurrence, but you can't just assume that it's always going to work out that way.

    So the final result is that the power series about x=0 sets up just this recurrence equation in the coefficients:

    $\displaystyle 5(k+2)(k+1) a_{k+2} + (k^2 - 2k + 10) a_k = 0$ for $\displaystyle k \ge 0$.

    (Notice that recurrence has a step of size 2... $\displaystyle a_{k+2}$ versus $\displaystyle a_k$, and so will be entirely determined by the two numbers $\displaystyle a_0, a_1$.

    But $\displaystyle y(0) = a_0, y'(0) = a_1$, so the power series solution of that 2nd order ODE is totally determined by two initial conditions, as expected.)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2015
    From
    US
    Posts
    6
    Thanks
    1

    Re: Seek power series solutions of the given differential equation about the given po

    @johnsomeoneThank you so much! This helps a lot. a quick question. So, when you use the peeling off method, say for example the highest initial value is k=2 as in my question that you went through. You would go through each series plugging in the starting k value until k=2 is reached? like for the first series it starts at k=0 so you plugged in k=0 and k=1?

    also another question, which I should probably know after you explaining a lot but, I understand now the recurrence relation but how would I get the answers for the first part of my question?
    Seek power series solutions of the given differential equation about the given point
    x0.
    y= a2k+2
    y=a2k+3 because these k's are not the same k's as in the recurrence relation...
    Last edited by hc23881; Aug 30th 2015 at 10:19 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2015
    From
    US
    Posts
    6
    Thanks
    1

    Re: Seek power series solutions of the given differential equation about the given po

    oh i can see the logic now. since my recurrence relation is ak+2 = then for y = a2k+2 = i set 2k = k

    and for a2k+3 i set k = 2k +1 and just plug in... but why is this the case?


    update: sorry to be asking so many questions... but when finding the first four terms for y1 and y2 how do you even get x's in your answers? I am just plugging in k=0,1,2,3 for my two equations..
    Last edited by hc23881; Aug 30th 2015 at 11:10 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Power series - Differential equation
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: Mar 26th 2011, 04:43 AM
  2. Finding differential equation from power series
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Feb 16th 2010, 11:11 PM
  3. Writing a differential equation in power series
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Dec 9th 2009, 01:48 AM
  4. Power series and differential equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 16th 2007, 12:00 PM
  5. Differential equation and power series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 11th 2007, 03:23 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum