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First Order Seires Solution ODE

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data
    y' = [tex]\sqrt{(1-y^2)
    }[/tex]
    Initial condition y(0) = 0
    a) Show y = sinx is a solution of the initial value problem.
    b) Look for a solution of the initial value problem in the form of a power series about x = 0. Find coefficients up to the term in x^3 in this series.

    2. Relevant equations

    part a) was Ok.

    3. The attempt at a solution
    This is for my part b attempt.

    (y')^2 + (y)^2 = 1

    (cosx)^2 + (sinx)^2 = 1

    [tex]\sum(-1)^ {2n} * (x^ {4n} )}/(2n!)^2[/tex] + [tex]\sum (-1)^{2n} * (x^ {4n+2})/((2n+1)!)^2[/tex] = 1

    I had also tried the general solution series for y and y'. y = a0 + a1x + ... + an*x^n
    y' = a1 + ... + n*an x^n
    Please note the values such as a0, a1, ...an have the nth term as a subscript.

    Maybe I should not be using part a where y=sinx and y'=cosx because I will have even powers when squared always...?

    What series will allow me to evaluate at the x^3 term?
     
  2. jcsd
  3. Nov 17, 2009 #2

    Mark44

    Staff: Mentor

    In your work for part b, you are assuming that you already have a solution. Start with y(x) = a_0 + a_1x +a_2x^2 + ... + a_nx^n + ...

    You don't need very many terms in the series above, since you need to work only up to the 3rd degree term in your solution.
     
  4. Nov 17, 2009 #3
    Hmm

    Let y = sinx be a solution of y' = sqrt(1-y^2)

    integral ( y') = integral {sqrt( 1 - sin^2(x) )}

    integral (dy) = intergral {cos(x) dx}

    y = sin(x)

    Given y(0) = 0

    We have that,
    y(0) = sin(0) = 0

    ==> y = sin(x) is a solution of the initial value problem

    Hence y = summation ( (-1)^n * x^ 2n+1 )/ (2n+1)!
    for the terms to x^3 we have that summation ( (-1)^n * x^ 2n+1 )/ (2n+1)! = x - x^3/6

    I know you're not my instructor would I be able to say from part a) we know that y = sinx is a solution Hence y = summation ( (-1)^n * x^ 2n+1 )/ (2n+1)!
    for the terms to x^3 we have that summation ( (-1)^n * x^ 2n+1 )/ (2n+1)! = x - x^3/6
     
  5. Nov 17, 2009 #4

    Mark44

    Staff: Mentor

    NO!
    In your first post it says
    "b) Look for a solution of the initial value problem in the form of a power series about x = 0."

    I already told you in my previous post what you need to do to accomplish that.
     
  6. Nov 17, 2009 #5
    y = n=0--> ∞ sum(a_n*x^n)

    y' = n=1-->∞ sum(n*a_n*x^n-1)

    y' = sqrt(1-y^2)

    y = sqrt(1-(y')^2)

    y(0) = 0

    sqrt(1-(y')^2) = 0

    (n=1 --> ∞ sum(n*a_n*x^n-1))^2 = 1

    Solved for y = 0

    Now I have sum for (y')^2 = 1

    I am having ambiguity with the general series for y(x) = a_0 + a_1x +a_2x^2 + ... + a_nx^n + ...
     
  7. Nov 18, 2009 #6

    Mark44

    Staff: Mentor

    1. Start with y(x) = a_0 + a_1x +a_2x^2. That should probably be enough terms.
    2. Calculate (y(x))^2
    3. Calculate 1 - (y(x))^2
    4. Calculate y'(x)
    5. Calculate (y'(x))^2
    Equate the expressions you got in the last two steps and find the coefficients up to the 3rd degree term. What you're doing in the last two steps is setting y'^2 = 1 - y^2, which seems easier than setting y' = sqrt(1 - y^2). The two are not equivalent, so it's possible you'll get a solution of y'^2 = 1 - y^2 that is not also a solution of your original differential equation.
     
  8. Nov 18, 2009 #7
    I believe that it is impossible to obtain a recursion series solution up to the x^3 term.

    However I did recognize that

    y' = sqrt(1-y^2)

    dy/dx = sqrt(1-y^2)

    dy/sqrt(1-y^2) = dx

    integral (dy/sqrt(1-y^2)) = integral (dx)

    inverse sin (y) = x

    y = sin(x)

    y(0) = 0

    y = sin(0) = 0

    Hence the series solution is y = sinx = series(sinx) = x - (x^3)/6 for values up to the x^3.
     
  9. Nov 18, 2009 #8

    Mark44

    Staff: Mentor

    In my opinion, that is NOT what you are supposed to do for part b. You are supposed to start with a power series, not find a solution and then write that as a power series. I have told you what I believe you need to do in posts #2, #4, and #6.
     
  10. Nov 19, 2009 #9
    I don't think that it is possible start with a series solution. The question says look for a solution of the initial value problem in the form of a power series. Not only can we can show that this is a solution of the initial value problem, but by using Taylor's expansion, sin(x) is in the form of a well known power series. Expanding the terms for a general summation of y an y' will not lead to a recursion formula, which is necessary to find a power series. This is sure to be the only way to find a solution.
     
  11. Nov 19, 2009 #10

    Mark44

    Staff: Mentor

    Apparently you don't.
    YES!! And that means that you should START with a GENERAL power series. The differential equation y' = sqrt(1 - y2) is very simple to solve by separation of variables, but I don't believe that is the intent of part b of this problem.
    Not so. You don't need a recursion formula - all you need are the terms up to degree three.
    I disagree - I have shown you a different approach, but have it your way, I'm done here. You asked for help, and I provided some help in the direction I believe this problem wants you to go. If you choose not to take the advice I offered, that's your choice.
     
  12. Nov 23, 2009 #11
    I have still been working with this problem. I've tried general series, squaring them, binomial expansions. Equating coefficients. This is IMPOSSIBLE to solve numerically.
     
  13. Nov 23, 2009 #12
    The homework has already been turned in, but I challenge you to prove me wrong and solve this numerically.

    http://www.math.pitt.edu/~troy/classes/math1270.html

    You can see that Homework 3 was due on the 18th
    The book was Boyce and DiPrima 9th Ed. Elementary Diff Eq and Boundary Value Problems.
     
  14. Nov 23, 2009 #13

    Mark44

    Staff: Mentor

    It's impossible the way you've been trying to do it, which is to approach it with a general series. It's not impossible if you take only a small number of terms in your series, say three at the most.
     
  15. Nov 23, 2009 #14
    I'm coming closer, but my results are still not precise.

    y = a_0 + a_1x + a_2x^2

    y(0) = 0

    y(0) = a_0 + 0 + 0 = 0 ==> a_0 = 0

    y = a_1x + a_2x^2
    y' = a_1 + 2a_2x

    (y)^2 = (a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4
    (y')^2 = (a_1 + 2a_2x)^2


    (y')^2 = 1 - (y)^2

    (a_1 + 2a_2x)^2 = 1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4}

    y' = 1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4} / {(a_1 + 2a_2x)}

    dy/dx = 1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4} / {(a_1 + 2a_2x)}


    INT [(a_1 + 2a_2x) dy] = INT[1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4}]

    a_1y + 2a_2xy = x - [(a_1)^2*x^3] /3

    y(a_1 + 2a_2x) = x - [(a_1)^2*x^3] /3

    I see it, but I'm stuck at this moment. Please help if I have a particular step incorrect.
     
  16. Nov 24, 2009 #15

    Mark44

    Staff: Mentor

    Looks fine up to here. Now move everything over to one side so that 0 is on the other side, and group together terms in the same power of x. What you'll have on the left side is a polynomial in x, and on the right side, another polynomial in x, all of whose coefficients are zero.

    Now equate the coefficients of 1, x, x^, and x^3 on both sides to solve for your a_1, a_2, and a_3. They will give you your coefficients for you Maclaurin series, which is the goal of this whole exercise.
     
  17. Nov 24, 2009 #16
    Are you interpreting a_3 as another coefficient in my work? I.e. a_1*a_2 = a_3
     
  18. Nov 24, 2009 #17

    Mark44

    Staff: Mentor

    No, I meant a_3 as the coefficient of x^3 in the series.
     
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