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

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

Question
Laplace transform
asked 2020-10-26
Find the inverse laplace transform of
\(\frac{6s+9}{s^2+17} s>0\)
\(y(t)=\dots\)

Answers (1)

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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