Whats the partial fraction for $f(x) = (-2s^2 -1)/(s^2(s^2+2))$

Question

Im trying to find the answer for the following inverse laplace, but when doing that $A/s$ + $B/s^2$ + $(Cs+D)/(s^2+2)$ , Im finding that b = -1/2 ; d= -3/2 but a and c are undefinied.

Answer 1

We have the system

$$\begin{align} x'' &= -x + y, x(0) = 0, x'(0) = -2 \\ y'' &= x - y, y(0) = 0, y'(0) = 1 \end{align}$$

Taking the Laplace transform of that system, we have

$$\begin{align} (s^2+1) X(s) - Y(s) &= -2 \\ - X(s) + (s^2+1) Y(s) &= 1 \end{align}$$

Multiplying the top equation by $(s^2+1)$ and adding it to the second equation

$$X(s) = -\dfrac{2s^2 + 1}{s^2(s^2+2)} = -\frac{3}{2 \left(s^2+2\right)}-\frac{1}{2 s^2}$$

The last expansion was done using

$$-\dfrac{2s^2 + 1}{s^2(s^2+2)} = \dfrac{a}{s} + \dfrac{b}{s^2} + \dfrac{cs + d}{s^2+2}$$

This reduces to

$$-2s^2-1 = 2b + (a + c)s^3 + (b + d)s^2 + 2 a s$$

Equating terms

$$2 a = 0, 2b = -1, b + d = -2, a + c = 0$$

This gves

$$a = c = 0, b = -\dfrac{1}{2}, d = -\dfrac{3}{2}$$

The solution should be

$$ x(t) = \dfrac{1}{4} \left(-2 t-3 \sqrt{2} \sin \left(\sqrt{2} t\right)\right) \\ y(t)= \dfrac{1}{4} \left(-2t+3 \sqrt{2} \sin \left(\sqrt{2} t\right)\right)$$