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Help needed in solving an IVP D.E. over 2 intervals

  1. Dec 13, 2008 #1
    1. The problem statement, all variables and given/known data

    y" +4y = g(x) ; y (0) = 1 & y(0) = 2

    g(x) =
    sinx , 0 [tex]\leq[/tex]x [tex]\leq[/tex] pi/2
    0 , x [tex]\succ[/tex] pi/2

    2. Relevant equations



    3. The attempt at a solution

    1> g(x) = sin x

    solving i get :
    yc = ec1sin2x + ec2cos2x

    US set of sin x = {sinx, cosx} so yp = Asinx + Bcosx

    solving i get:
    yp = (1/3)sinx

    so
    y = ec1sin2x + ec2cos2x + (1/3)sinx ...(1)

    for the conditions given i get:

    c1 = 5/6e & ec2 = 1/e

    thus (1) becomes:
    y = (5/6)sin2x + cos2x + (1/3)sinx .... (2) for x between 0 & pi/2


    2> for g(x) = 0 with x greater than pi/2

    we have : y" + 4y = 0
    with m1 = 2i & m2 = -2i

    y = ec1sin2x +ec2cos2x ....(3)

    solving for the conditions giver we get:

    c1 = 1/ e = c2 thus
    y = sin 2x + cos 2x ..... (4)

    from here on i can;t figure how to continue to find a solution / y & y' are continuous @ x = pi/2

    i need help plz.

    note that the exercise is long enough to post all the details of the attempt of solution. i hope u reply. thx
     
  2. jcsd
  3. Dec 14, 2008 #2

    tiny-tim

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    Hi bobmerhebi! :smile:

    (have a ≤ and a ≥ and a π :wink:)

    (and why are you using the difficult-to-read-and-type c1 and c2 instead of the usual A and B? :rolleyes:)

    (and what is e?)

    Very good …

    now just find y(π/2) and y'(π/2), and use those as your new initial conditions. :smile:
     
  4. Dec 14, 2008 #3
    ur right about c1 & c2 but i got used to use them for the homog. solution & A & B... for the particulat sol.'s

    after calculating for [tex]\pi[/tex]/2 for both y & y' with x btw 0 & [tex]\pi[/tex]/2 i get:

    y([tex]\pi[/tex]/2) = -2/3 &
    y'([tex]\pi[/tex]/2) = -1 . but now what?

    im asked to find a solution so that y & y' are continuous @ [tex]\pi[/tex]/2.
     
  5. Dec 14, 2008 #4

    tiny-tim

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    Hi bobmerhebi! :smile:

    (what happened to that π i gave you? :rolleyes:)

    Isn't y'(π/2)-= -5/3?

    Anyway, just solve the particular solution all over again, with y(π/2) and y'(π/2) as your new boundary conditions. :smile:
     
  6. Dec 14, 2008 #5
    sorry but i still dont get it.

    lets go all over again.
    1st u asked what is e. its the exponential. why ? is there anything wrong with it?

    2nd: u told me to take a = [tex]\pi[/tex], less than & grater than [tex]\pi[/tex]. why is that? & what do u mean by a?

    then u said to use y([tex]\pi[/tex]/2) & y'([tex]\pi[/tex]/2). but which y? the one with g(x) = sin x or 0?

    as for the y'([tex]\pi[/tex]/2). if its y' of y with x btw 0 & [tex]\pi[/tex]/2 then y'([tex]\pi[/tex]) = (5/6)cos[tex]\pi[/tex] -2sin[tex]\pi[/tex] +(1/3)sin[tex]\pi[/tex]/2 = -4/3

    finally ur saying to solve y_p again for those values. how is that if i already have the coefficients?

    thx
     
  7. Dec 14, 2008 #6
    sorry but i still dont get it.

    lets go all over again.
    1st u asked what is e. its the exponential. why ? is there anything wrong with it?

    2nd: u told me to take a = [tex]\pi[/tex], less than & grater than [tex]\pi[/tex]. why is that? & what do u mean by a?

    then u said to use y([tex]\pi[/tex]/2) & y'([tex]\pi[/tex]/2). but which y? the one with g(x) = sin x or 0?

    as for the y'([tex]\pi[/tex]/2). if its y' of y with x btw 0 & [tex]\pi[/tex]/2 then y'([tex]\pi[/tex]) = (5/6)cos[tex]\pi[/tex] -2sin[tex]\pi[/tex] +(1/3)sin[tex]\pi[/tex]/2 = -4/3

    finally ur saying to solve y_p again for those values. how is that if i already have the coefficients?

    thx
     
  8. Dec 14, 2008 #7

    tiny-tim

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    Yes! Why is it there? It's multiplied by an arbitrary constant anyway, so ec1 or ec2 is just the same as c1 or c2.

    erm … look at the signature at the bottom of this post! :biggrin:

    ah, but isn't it (1/3)cosπ/2?
    g(x) = 0, because that's for the interval you'll be dealing with. :smile:
     
  9. Dec 14, 2008 #8
    y = (5/6)sin2x + cos2x + (1/3)sinx .... (2) with 0 ≤ x ≤ π/2

    y = sin 2x + cos 2x ..... (4) with x (just greater than not or equal) π/2

    & y(π/2) = -2/3 & y'(π/2) = -5/3

    then i should use these new initial conditions with (4). for what? i have y = sin 2x + cos 2x ..... (4)

    sorry, it might be so simple but im tired & i have a culture exam 2moro & a 25 page assignment that;s still unfinished. so excuse.
     
  10. Dec 14, 2008 #9

    tiny-tim

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    Sorry … I'm not seeing where the problem is …

    you worked out the ≤ π/2 section fine :smile:

    just go through the same procedure for the ≥ π/2 section …

    by definition, they'll fit together continuously. :wink:

    Get some sleep now :zzz: and try it again in the morning :smile:
     
  11. Dec 14, 2008 #10
    thx i will retry it late this night. unfortunately i have to submit my assignment 2moro morning. but never mind. thx for ur sincere help
     
  12. Dec 14, 2008 #11
    thx i will retry it late this night. unfortunately i have to submit my assignment 2moro morning. but never mind. thx for ur sincere help
     
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