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General solution to 2nd order DE

  1. Oct 31, 2009 #1
    Hi, im just looking for clarification whether or not im doing these problems correctly, heres an example:

    1. The problem statement, all variables and given/known data
    Find the general solution to
    x'' -2x' + 5x = 0


    2. Relevant equations
    Charasteristic polynomial.
    Quadratic equation.
    General solution form for complex roots.

    3. The attempt at a solution
    form the char polynomial:
    m^2 -2m +5 = 0

    Solve for roots:
    (-2 (+-)SQRT(2^2 - 4(1)(5)))/2
    -1+SQRT(-16)
    -1-SQRT(-16)
    m1 = -1+4i
    m2 = -1-4i

    Insert into general solution form:

    y = (e^ax)(Acoswx + Bsinwx)
    y = (e^-1x)(Acos4ix + Bsin4ix)

    General solution: (?)
    y = (e^-x)(Acos4ix + Bsin4ix)
     
  2. jcsd
  3. Oct 31, 2009 #2

    Mark44

    Staff: Mentor

    Almost!
    Your general solution is
    y = e-x(Acos(4x) + Bsin(4x))

    You can verify that this is the solution by calculating y' and y'' and substituting them into your differential equation.
     
  4. Oct 31, 2009 #3
    Nice, thanks!
     
  5. Nov 2, 2009 #4
    What do I do if im asked to write a solution that satisfies the conditions
    x(0) = 1
    x´(0) =0
     
  6. Nov 2, 2009 #5

    Mark44

    Staff: Mentor

    Your solution is actually x(t) = e-t(Acos(4t) + Bsin(4t)). I hadn't noticed earlier that you had switched your solution to be y as a function of x.

    Use your initial conditions to substitute x = 1 and t = 0 into your solution, and x' = 0 and t = 1 into the derivative of your solution. That will give you two equations in two unknowns, so you should be able to solve for the constants A and B.
     
  7. Nov 2, 2009 #6
    Cool, I got it now, thx.
     
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