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Inverse Laplace transform for an irreducible quadratic?

  1. May 3, 2017 #1
    1. The problem statement, all variables and given/known data
    I have to take the inverse Laplace of this function (xoms+bxo)/(ms2+bs+k) this can not be broken into partial fractions because it just gives me the same thing I started with. How is this done? This is coming from the laplace of the position function for a harmonic oscillator with initial conditions x(o) = xo and dx/dt(0) = 0 if that helps. The original function is mx''+bx'+kx = 0 where m is mass, b is coefficient of damping, k is spring constant. The end goal of this whole thing is to solve this ODE using laplace.

    2. Relevant equations
    mx''+bx'+kx = 0

    3. The attempt at a solution
    I have took the Laplace of the above equation and got down to the point where I now need to take the inverse Laplace of (xoms+bxo)/(ms2+bs+k)
     
  2. jcsd
  3. May 3, 2017 #2

    Ray Vickson

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    Let the two roots of ##m s^2 + bs + k=0## be ##r_1,r_2##. If ##r_1 \neq r_2## you can always write ##1/(m s^2 + bs + k)## as the partial fraction
    $$\frac{1}{m} \left( \frac{A}{s-r_1} + \frac{B}{s - r_2} \right), $$
    and easily enough find ##A,B##. If ##r_1 = r_2 = r## the partial fraction expansion of ##f(s) = (1/m) 1/(s-r)^2## is just ##f(s)## itself.

    So, when ##r_1 \neq r_2,## you need to invert
    $$\frac{A}{m} \frac{us+v}{s-r_1} + \frac{B}{m} \frac{us+v}{s-r_2},$$
    which is easy enough. When ##r_1 = r_2 = r## you need to invert
    $$\frac{1}{m} \frac{u s + v}{(s-r)^2},$$
    which is pretty standard and can be found in tables, etc.
     
  4. May 3, 2017 #3
    You should show your work to see if anyone can make some suggestions or spot any errors.

    I would start by dividing numerator and denominator by m. It makes it cleaner to solve.
     
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