Results 1 to 10 of 10
Like Tree3Thanks
  • 2 Post By Archie
  • 1 Post By hc23881

Thread: Consider the following. y'' − xy' − y = 0

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

    Consider the following. y'' − xy' − y = 0

    Consider the following.y''xy'y = 0, y(0) = 4, y'(0) = 1

    (a) Find the first five nonzero terms in the solution of the given initial value problem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,493
    Thanks
    899
    Awards
    1

    Re: Consider the following. y'' − xy' − y = 0

    How far has you been able to get?

    Start with
    $\displaystyle y = \sum_{n = 0}^{\infty}a_n x^n$

    $\displaystyle y' = \sum_{n = 1}^{\infty}n~a_n x^{n-1}$

    $\displaystyle y'' = \sum_{n = 2}^{\infty}n(n - 1)~a_n x^{n-2}$

    So your equation becomes
    $\displaystyle \sum_{n = 2}^{\infty}n(n - 1)~a_n x^{n-2} - x \cdot \sum_{n = 1}^{\infty}n~a_n x^{n-1} - \sum_{n = 0}^{\infty}a_n x^n = 0$

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2013
    From
    Colombia
    Posts
    1,993
    Thanks
    727

    Re: Consider the following. y'' − xy' − y = 0

    I would note that the (MacLaurin) series expansion of the solution $y(x)$ about $x=0$ is
    $$y(x) = y(0) + {y'(0) \over 1!}x + {y''(0) \over 2!}x^2 + \cdots + {y^{(n)}(0) \over n!}x^n + \cdots$$
    The initial conditions give us $y(0)$ and $y'(0)$ and the equation gives us
    $$y''(0) = 0\cdot y'(0) + y(0) = y(0)$$
    We can then differentiate (implicitly) the original equation to get
    $$y''' - xy'' - 2y' = 0 \implies y'''(0) = 0\cdot y''(0) + 2y'(0)$$
    And repeat...
    Thanks from topsquark and romsek
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,249
    Thanks
    3366

    Re: Consider the following. y'' − xy' − y = 0

    hc23881, is there a reason why you made no attempt to do the problem yourself? If you know the basic definitions, this problem is a little tedious but straightforward.
    Follow Math Help Forum on Facebook and Google+

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

    Re: Consider the following. y'' − xy' − y = 0

    Consider the following. y'' − xy' − y = 0-solution.png

    I know those are teh solutions but im not sure about using the given initial values..? does y(o) equal to a0? and y'(o) = a1?

    edit:i was able to get the answer
    Last edited by hc23881; Sep 2nd 2015 at 10:58 AM.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,249
    Thanks
    3366

    Re: Consider the following. y'' − xy' − y = 0

    So this is just a question of arithmetic? If $\displaystyle y(x)= a_0+ a_1x+ a_2x^2+ a_3x^3+ \cdot\cdot\cdot$ where all of the terms in the "$\displaystyle \cdot\cdot\cdot$" involve positive powers of x. So $\displaystyle y(0)= a_0+ a_1(0)+ a_2(0)^2+ a_3(0)^3+ \cdot\cdot\cdot$. That is [itex]a_0= 0[/itex]. Similarly, differentiating that sum (it is a power series so differentiable term by term), we get $\displaystyle y'(x)= a_1+ 2a_2x+ 3a_3x^2+ 4a_4x^3+ \cdot\cdot\cdot$. So $\displaystyle y'(0)= a_1+ a_2(0)+ a_3(0)^2+ a_4(0)^3+ \cdot\cdot\cdot$.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2015
    From
    US
    Posts
    6
    Thanks
    1

    Re: Consider the following. y'' − xy' − y = 0

    Thanks I got that one! I am now having trouble on part of this one
    (5 + x2)y'' − xy' + 10y = 0, x0 = 0

    I was able to find the recurrence relation: an+2 = -(n^2 -2n+10)/(5(n+2)(n+1))
    but I cant get this part
    Find the first four terms in each of two solutions
    y1
    and
    y2
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2013
    From
    Colombia
    Posts
    1,993
    Thanks
    727

    Re: Consider the following. y'' − xy' − y = 0

    $x_0 = 0$? That's just shorthand to say that we want the series expansions about $x=0$?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2013
    From
    Colombia
    Posts
    1,993
    Thanks
    727

    Re: Consider the following. y'' − xy' − y = 0

    $x_0 = 0$? That's just shorthand to say that we want the series expansions about $x=0$?

    You don't need a recurrence relation (which that isn't because you only have $a_{n+2}$).

    Instead look to find expressions for $a_1$ and $a_2$ in terms of $a_0$ and $a_1$.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,249
    Thanks
    3366

    Re: Consider the following. y'' − xy' − y = 0

    Quote Originally Posted by hc23881 View Post
    Thanks I got that one! I am now having trouble on part of this one
    (5 + x2)y'' − xy' + 10y = 0, x0 = 0

    I was able to find the recurrence relation: an+2 = -(n^2 -2n+10)/(5(n+2)(n+1))
    This is NOT a "recurrence relation". Did you mean an+2 = -(n^2 -2n+10)/(5(n+2)(n+1))an?

    but I cant get this part
    Find the first four terms in each of two solutions
    y1

    y2
    The simplest way to handle this is to assume y1(0)= 1, y1'(0)= 0 and y2(0)= 0, y2(0)= 1.
    That will guarantee two independent solutions.

    Since y(0)= a0 and y'(0)= a1, you can see immediately that, for y1, a0= 1 and a1= 0. For y2, a0= 0 and a1= 1. Further, since (assuming I was correct above) each an is a multiple of an-2, it is clear that, for y1, all an with n odd is 0, and, for y2, all n with n odd is 0. y1 will be a sum with only even powers of x and y2 will be a sum with only odd powers of x.

    Now, taking a0= 1 for y1, what is a2? Using that, what is a4?

    Taking a1= 1 for y2, what is a3? Using that, what is a5.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Dec 27th 2014, 11:35 PM
  2. Replies: 3
    Last Post: Aug 6th 2014, 06:38 AM
  3. Replies: 1
    Last Post: Apr 20th 2013, 05:46 PM
  4. Solve y ′′ − 3 y ′ − 4 y = −3 x+5
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Jun 2nd 2010, 11:50 PM
  5. Replies: 8
    Last Post: Apr 19th 2009, 02:41 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum