# Inverse Laplace transform for an irreducible quadratic?

1. May 3, 2017

### Vitani11

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. May 3, 2017

### Ray Vickson

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.

3. May 3, 2017

### magoo

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.