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2nd order homogeneous diff eq

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data
    y"-2y'+5=0, y(∏/2)=0, y'(∏/2)=2

    find general solution of this diff eq

    2. Relevant equations



    3. The attempt at a solution

    i have followed all of the steps for this, rather easy 2nd order diff eq, but i my solution differs from the books solution.

    steps:
    r2-2r+5=0

    r 1/2= 1±2i

    y=c1etcos(2t)+c2etsin(2t)
    we have y(∏/2)=0, when replaced in eq above gives me
    c1=-e-(∏/2)

    calculating y' from the above y gives:
    y'=et((c2-2c1)sin(2t)+(c1+2c2)cos(2t))
    replacing y'(∏/2)=2 i get c2=(3/2)e-(∏/2)

    putting back c1 and c2 i get final solution of y=(3/2)e(t-∏/2)sin(2t)-e(t-∏/2)cos(2t)
    book saying that the solution of this eq is y=-e(t-∏/2)sin(2t)

    can someone point out where i am making mistakes, since i did this problem few times and i always get the same answer....
     
  2. jcsd
  3. Oct 6, 2012 #2

    vela

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    This actually gives you ##y(\pi/2)=0=c_1e^{\pi/2}##. You set it equal to 1, not 0.

     
  4. Oct 6, 2012 #3
    vela,
    thank you.. i knew i made some silly mistake...

    after correcting my work i still have discrepancy in sign.. i am getting positiv, while book is saying negative..

    After differentiating y I get => y'=et((c2-2c1)sin(2t)+(c1+2c2)cos(2t))

    since c1=0 and y'(∏/2)=2

    2=2e∏/2c2 which is

    c2=e-(∏/2) and i need this to be negative......
     
  5. Oct 6, 2012 #4
    got it.. cos(pi)=-1 and there i get my negative sign... Vela, thanks alot for pointing my mistake
     
  6. Oct 6, 2012 #5

    vela

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    Glad you figured it out.
     
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