Inverse Laplace transform for an irreducible quadratic?

[Vitani11]

Homework Statement

I have to take the inverse Laplace of this function (x_oms+bx_o)/(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) = x_o 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.

Homework Equations

mx^''+bx^'+kx = 0

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 (x_oms+bx_o)/(ms²+bs+k)

 

[Ray Vickson]

Homework Statement

I have to take the inverse Laplace of this function (x_oms+bx_o)/(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) = x_o 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.

Homework Equations

mx^''+bx^'+kx = 0

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 (x_oms+bx_o)/(ms²+bs+k)

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.

 

[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.