Consider the following problem: determine $\displaystyle\mathscr{L}^{-1}\bigg[\frac{s^2+1}{s^2(s^2-4s+9)}\bigg]$ using formulas and using convolutions.
Using the formulas I found that the solution would be: $\displaystyle\frac{1}{81}\cdot\Big[9t+16\sqrt{5}e^{2t}sin(\sqrt{5}t)-4e^{2t}cos(\sqrt{5}t)+4\Big].$
But using convolutions we encountered problems. I thus considered: $H(s)=\displaystyle\frac{s^2+1}{s^2(s^2-4s+9)}, F(s)=\displaystyle\frac{s^2+1}{s^2}$ and $G(s)=\displaystyle\frac{1}{s^2-4s+9}\Rightarrow f(t)=\delta_t+t$ and $g(t)=\displaystyle\frac{1}{\sqrt{5}}e^{2t}sin(\sqrt{5}t)$ so $\displaystyle h(t)=f(t)*g(t)=\int_{0}^tf(t-\tau)g(\tau)d\tau=\int_{0}^{t}\Big(\delta(t-\tau)+t-\tau\Big)\cdot\Big(\frac{1}{\sqrt{5}}e^{2\tau}sin(\sqrt{5}\tau)\Big)d\tau.$
At this point things are getting messy for me because I don't know what I can do with the Dirac function right now and what property to use for it.
If possible a hint to continue. Thank you
Best Answer
I think it is easier to do it by partial fractions, but since you want to use convolution I will try to do it in the best way:
Let $F(s)=\dfrac{s^2+1}{s^2(s-2)}=\dfrac{-1/2}{s^2}+\dfrac{-1/4}{s}+\dfrac{5/2}{s-2}$ and $G(s)=\dfrac{s-2}{(s-2)^2+5}$. Then $f(t)=-\dfrac{t}{2}-\dfrac{u(t)}{4}+\frac{5}{4}e^{2t}$ and $g(t)=e^{2t}\cos(\sqrt 5t)$. Since $\mathscr{L}^{-1}(F(s)G(s))=(f*g)(t)$, then $$ (f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau=\int_0^t \left(-\dfrac{\tau}{2}-\dfrac{\overbrace{u(\tau)}^{1}}{4}+\frac{5}{4}e^{2\tau}\right)e^{2t-2\tau}\cos(\sqrt 5t-\sqrt 5\tau)d\tau $$ You can continue with the process, through integration by parts.
Doing the math, the following remains: $$ \int \left(-\dfrac{\tau}{2}-\dfrac{1}{4}+\frac{5}{4}e^{2\tau}\right)e^{2t-2\tau}\cos(\sqrt 5t-\sqrt 5\tau)d\tau=e^{2t-2\tau}\left(\frac{(9x+4)}{81}\cos(\sqrt 5t-\sqrt 5\tau)+\frac{\sqrt 5}{324}(18x+17)\sin(\sqrt 5t-\sqrt 5\tau)-\frac{\sqrt 5}{4}e^{2\tau}\sin(\sqrt 5t-\sqrt 5\tau)\right)+C $$ Doing the accounts again you arrive at the same thing as using partial fractions.