How to find inverse laplace transform
$$ F(s) = \dfrac{6s+9}{s^2-10s+29} $$
How do you solve the inverse Laplace transform of this above equation?
$\endgroup$ 53 Answers
$\begingroup$$$ F(s) = \dfrac{6s+9}{s^2-10s+29}=\dfrac{6s-30+39}{(s-5)^2+4}=\dfrac{6s-30}{(s-5)^2+4}+\dfrac{39}{(s-5)^2+4}=6\dfrac{s-5}{(s-5)^2+4}+\dfrac{39}{2}\dfrac{2}{(s-5)^2+4} $$ Then $$ \mathcal{L}^{-1}\{F(s)\} =6e^{5t}\cos{2t}+\frac{39}{2}e^{5t}\sin{2t} $$
$\endgroup$ $\begingroup$We have $s^2-10s+29=(s-5)^2+2^2$, which implies that $$F(s)=\frac{6s+9}{s^2-10s+29}=\frac{6(s-5)+39}{(s-5)^2+2^2}.$$ Therefore, $$\mathcal{L}^{-1}(F(s))=6\mathcal{L}^{-1}\left(\frac{s-5}{(s-5)^2+2^2}\right) +\frac{39}{2}\mathcal{L}^{-1}\left(\frac{2}{(s-5)^2+2^2}\right)\\ =6e^{5t}\cos(5t)+\frac{39}{2}e^{5t}\sin(2t)$$ by here.
$\endgroup$ $\begingroup$$$ \underbrace{s^2-10s+29 = (s-5)^2+2^2}_{\text{completing the square}} = t^2+2^2 $$ So $$ \frac{6s+9}{s^2-10s+29} = \frac{6(s-5)+39}{(s-5)^2+2^2} = \frac{6t+39}{t^2+2^2} = 6\frac{t}{t^2+2^2} + \frac{39}{2}\cdot\frac{2}{t^2+2^2} $$ Now look at each term separately.
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