# Find Laplace transform for frac{s+8}{s^2+4s+5}

Question
Laplace transform
Find Laplace transform for
$$\frac{s+8}{s^2+4s+5}$$

2021-02-01

Step 1
Formula Used:
$$L\left\{e^{at}\sin bt\right\}=\frac{b}{(s-a)^{2}+b^{2}}$$
$$L\left\{e^{at}\cos bt\right\}=\frac{s-a}{(s-a)^{2}+b^{2}}$$
Step 2
$$\text{Consider }F(s)=\frac{s+8}{s^2+5s+5}$$
$$=\frac{s+8}{(s+2)^{2}+1}$$
$$=\frac{s+2+6}{(s+2)^{2}+1}$$
$$=\frac{(s+2)}{(s+2)^{2}+1}+\frac{6}{(s+2)^2+1}$$
Step 3
$$\text{Consider }F(s)=\frac{(s+2)}{(s+2)^{2}+1}+\frac{6}{(s+2)^2+1}$$
We now apply Inverse Laplace Transform, we get
$$L^{-1}\left\{F(s)\right\}=L^{-1}\left\{\frac{(s+2)}{(s+2)^2+1}\right\}+6L^{-1}\left\{\frac{1}{(s+2)^2+1}\right\}$$
$$\Rightarrow\ f(t)=e^{-2t}\cos t+6e^{-2t}\sin t$$
$$=e^{-2t}(\cos t+6\sin t)$$
Step 4
: The inverse Laplace Transform of F(s) is
$$f(t)=e^{-2t}(\cos t+6\sin t)$$

### Relevant Questions

Use Laplace transform to evaluate the integral
$$\int_0^\infty te^{-2t} \sin(2t)dt$$
a) $$\frac{1}{B}$$
b) not defined
c) $$\frac{4s}{(s^2+4)^2}$$
d) $$\frac{4(s+2)}{(s^2+4s+8)^2}$$
determine a function f(t) that has the given Laplace transform
$$F(s)=\frac{4s+5}{s^{2}+9}$$
find the inverse Laplace transform of the given function
$$F(s)=\frac{e^{-2}+e^{-2s}-e^{-3s}-e^{-4s}}{s}$$
Write down the qualitative form of the inverse Laplace transform of the following function. For each question first write down the poles of the function , X(s)
a) $$X(s)=\frac{s+1}{(s+2)(s^2+2s+2)(s^2+4)}$$
b) $$X(s)=\frac{1}{(2s^2+8s+20)(s^2+2s+2)(s+8)}$$
c) $$X(s)=\frac{1}{s^2(s^2+2s+5)(s+3)}$$
Find inverse Laplace transform $$L^{-1}\left\{\frac{s-5}{s^2+5s-24}\right\}$$ Please provide supporting details for your answer
The inverse Laplace transform for
$$\displaystyle{F}{\left({s}\right)}=\frac{8}{{{s}+{9}}}-\frac{6}{{{s}^{2}-\sqrt{{3}}}}$$ is
a) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \sin{{h}}{{\left({3}{t}\right)}}$$
b) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
c) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \sin{{h}}{\left({3}{t}\right)}$$
d) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
Solve for the Inverse Laplace Transformations. Show the solution.
$$F(s)=\frac{6}{(s^2+4s+20)^2}$$
Use properties of the Laplace transform to answer the following
(a) If $$f(t)=(t+5)^2+t^2e^{5t}$$, find the Laplace transform,$$L[f(t)] = F(s)$$.
(b) If $$f(t) = 2e^{-t}\cos(3t+\frac{\pi}{4})$$, find the Laplace transform, $$L[f(t)] = F(s)$$. HINT:
$$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)$$
(c) If $$F(s) = \frac{7s^2-37s+64}{s(s^2-8s+16)}$$ find the inverse Laplace transform, $$L^{-1}|F(s)| = f(t)$$
(d) If $$F(s) = e^{-7s}(\frac{1}{s}+\frac{s}{s^2+1})$$ , find the inverse Laplace transform, $$L^{-1}[F(s)] = f(t)$$
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$
$$x"+8x'+20x=f(t)$$
$$x(0)=-3$$
$$x'(0)=5$$
$$\text{where the forcing } f(t) \text{ is given by }$$
$$f(t) = \begin{cases} t^2 & \quad \text{for } 0\leq t<2 ,\\ 4e^{2-t} & \quad \text{for } 2\leq t < \infty . \end{cases}$$