Find the inverse laplace transform of frac{6s+9}{s^2+17} s>0 y(t)=dots

Question
Laplace transform
Find the inverse laplace transform of
$$\frac{6s+9}{s^2+17} s>0$$
$$y(t)=\dots$$

2020-10-27
Step 1
Given the Laplace transform of a function y(t) such that
$$L\left\{y(t)\right\}=\frac{6s+9}{s^2+17}$$
Find f(t) using inverse Laplace transform.
Step 2
First, separate the numerator.
$$\frac{6s+9}{(s^2+17)}=6 \frac{s}{(s^2+17)}+9 \frac{1}{(s^2+17)}$$
Since $$L^{-1}\left\{\frac{a}{(s^2+a^2)}\right\}=\sin (at)$$ , multiply the numerator and denominator on the second sum by $$\sqrt{17}$$
$$\frac{6s+9}{(s^2+17)}=6 \frac{s}{(s^2+(\sqrt{17})^2}+\frac{9}{(\sqrt{17})} \cdot \frac{\sqrt{17}}{s^2+(\sqrt{17})^2}$$
Step 3
Now, take inverse Laplace transform on both sides.
$$L^-1\left\{\frac{6s+9}{(s^2+17)}\right\}=6L^ {-1}\left\{\frac{s}{s^2+(\sqrt{17})^2}\right\}+\frac{9}{\sqrt{17}}L^-1\left\{\frac{\sqrt{17}}{s^2+(\sqrt{17})^2}\right\}$$
Since $$L^-1\left\{\frac{a}{(s^2+a^2)}\right\}=\sin (at) \text{ and } L^-1\left\{s/(s^2+a^2)\right\}=\cos(at)$$
$$L^-1\left\{\frac{6s+9}{(s^2+17)}\right\}=6\cos(\sqrt{17t})+9/(\sqrt{17})sin(\sqrt{17t})$$
Therefore, $$y(t)=6\cos(\sqrt{17t})+\frac{9}{(\sqrt{17})}\sin(\sqrt{17t})$$

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