Question

Laplace inverse
(3s+4)/(s^3-2s^2 +7s-14)

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Answer 1

Factoring the denominator gives

`s^3-2s^2+7s-14=s^2(s-2)+7(s-2)=(s-2)(s^2+7)`

(Keep in mind that
`7=(\sqrt7)^2`.)

Split up the given transform into partial fractions:

`(3s+4)/((s-2)(s^2+7))=(c_1)/(s-2)+(c_2s+c_3)/(s^2+7)`

`3s+4=c_1(s^2+7)+(c_2s+c_3)(s-2)`

`3s+4=(c_1+c_2)s^2+(-2c_2+c_3)s+7c_1-2c_3`

`\implies\begin{cases}c_1+c_2=0\\-2c_2+c_3=3\\7c_1-2c_3=4\end{cases}\implies c_1=(10)/(11),c_2=-(10)/(11),c_3=(13)/(11)`

`\implies(3s+4)/((s-2)(s^2+7))=\frac1{11}\left((10)/(s-2)-(10s-13)/(s^2+7)\right)`

Recalling the frequency shift property,

`F(s-a)=L(e^(at)f(t))`

we get

`F(s-2)=(10)/(s-2)\implies f(t)=10e^(-2t)L^(-1)\left(\frac1s\right)`

`\implies f(t)=10e^(-2t)`

The remaining inverse transforms reduce to sines and cosines:

`F(s)=(10s-13)/(s^2+7)=(10s)/(s^2+7)-(13)/(\sqrt7)(\sqrt7)/(s^2+7)`

`\implies f(t)=10\cos(\sqrt7\,t)-(13)/(\sqrt7)\sin(\sqrt7\,t)`

So we end up with

`L^(-1)\left((3s+4)/((s-2)(s^2+7))\right)=\boxed{\frac1{11}\left(10e^(-2t)-10\cos(\sqrt7\,t)+(13)/(\sqrt7)\sin(\sqrt7\,t)\right)\right)}`