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Initial value problem differentials, close to getting answer

  1. Jun 25, 2008 #1
    1. The problem statement, all variables and given/known data
    y" + 2y' + 10y=0
    y'(0)= -1

    solve initial value problem
    2. Relevant equations

    e^( ~ + iu)t= e^ ~t (cos ut + i sin ut)

    3. The attempt at a solution

    i've gotten pretty far into the problem, but i just can't seem to get the correct final answer.

    I changed y" + 2y' + 10y=0 int r^2 +2r +10= 0 and used the quadratic formula to get:

    r= 1-3i and r= 1+ 3i

    so for the general solution i got:
    yg= y1 + y2
    yg= c1e^t( cos 3t + i sin 3t) + c2e^t( cos 3t - i sin 3t)

    i need to solve for the constants:

    y= c1 e^t cos 3t + c2 i sin3t ( an example in the book followed this method but i dont know why there is a c2 next to sin 3t and why there isn't a y2 )

    i got c1= 1 and took the derivative to find c2 and got c2= -2/3

    y'= c1 e^t cos 3t -3c1e^t sin 3t + c2e^tsin3t + 3 cos3tc2e^t

    overall i got yg= 2e^tcos 3t but the answer is suppose to be yg= e^-t cos 3t. why? the textbook i use has this annoying ability of skipping 90% of the arithmatic work within examples, so i'm not sure where I could have gone wrong, can u guys help? :/
  2. jcsd
  3. Jun 25, 2008 #2


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    Homework Helper

    remember the quadratic eq'n formula is

    [tex]r=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

    you just forgot the '-b'
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