Differential Equation Help

Discussion in 'Math' started by Sparky, Jun 19, 2007.

  1. Sparky

    Thread Starter AAC Fanatic!

    Aug 1, 2005
    75
    0
    The problem is: Solve : y'' -y = cosh(x)

    I've attempted a solution and is attached.

    The answer in the back of the book agree with part of mine. However, the circled part - the part with 1/8 as a coefficient on e^x terms. It's circled on the bottom.

    Either the back of the book is wrong (and I'm correct) or this term should cancel and I've missed something.
    Help?

    Thanks
    -Sparky_
     
  2. AlexK

    Active Member

    May 23, 2007
    34
    0
    I didn't go through all your solution so i dont know if got it wrong or not.
    this is how i solved it:
    (1): y'' -y = cosh(x)=(exp(x)-exp(-x)/2

    y=yh+yp;

    yh=c*exp(-x)+c2exp(x);

    then we guess a private solution: yp=k1*x*exp(x)+k2*x*exp(-x);
    then we substitute this expression into (1) and find the constants k1,k2.
    So the final solution is: y=yh+yp;

    I solved it, and it matched the answer you said that was in the book.
    Hope i helped.
     
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