I got this equation by doing a Laplace transformation. Now, I want to find out the inverse Laplace and for that first I need to decompose this equation but I'm bit confused about how to express this equation as partial fraction. Moreover, the numerator of the second term(right side) is constant, do I need to decompose this also?
$$Y(s) = \frac{s+1}{s^2(s^2-2s+5)}+\frac4{s^2-2s+5}$$
leave the $\frac {4}{s^2 - 2s + 5}$ term alone while you break up $\frac {s+1}{s^2(s^2 - 2s + 5)}$
$\frac {s+1}{s^2(s^2 - 2s + 5)} = \frac {As+ B}{s^2} + \frac {Cs + D}{s^2 -2s + 5}$
$s +1 = (As + B)(s^2 -2s + 5) + Cs^3 + Ds^2$ $s +1 = (A+C)s^3 +(-2A + B+D)s^2 + (5A-2B)s + 5B$
$B = \frac 15$
$5A - \frac 25 = 1\\A = \frac {7}{25}$
$-\frac {14}{25} + \frac {1}{5} + D = 0\\ D = \frac {9}{25}$
$C = -\frac 7{25}$
$\frac {7}{25}\frac {1}{s} + \frac {1}{5}\frac {1}{s^2} -\frac {7}{25} \frac {s}{(s-1)^2 + 4} + \frac {9}{25}\frac {1}{(s-1)^2 + 4}+4\frac {1}{(s-1)^2 + 4}$
We can combine the $\frac {9}{25}$ with the $4.$
But to do our inverse tranform we really want terms to be in the form, $A\frac {s-c}{(s-c)^2 + a^2} + B\frac {a}{(s-c)^2 + a^2}$
$\frac {7}{25}\frac {1}{s} + \frac {1}{5}\frac {1}{s^2} -\frac {7}{25} \frac {(s-1)}{(s-1)^2 + 4} + \frac {51}{25}\frac {2}{(s-1)^2 + 4}$