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inverse laplase
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\[\frac{ 2s-5 }{ s ^{2} -4s +13 }\]
can you factor the denominator using 'completing the square' method
i have done the completing the square (s-2)+9
dont forget the ^2
sorry i forgot about that.... then i dont know the next step
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something like this \[(s-2)^2+3^2\]
so we have \[\mathcal L^{-1}\left\{\frac{ 2s-5 }{ s ^{2} -4s +13 }\right\}\] \[=\mathcal L^{-1}\left\{\frac{ 2s}{ (s-2)^2+3^2} -\frac{5 }{ (s-2)^2+3^2 }\right\}\] now apply the shift theorem
for 2s/(s-2)^2+9 whish teorem i shpuld appllied
Shift theorem\[\large\boxed{\mathcal L^{-1}\left\{F(s+a)\right\}=e^{-at}\mathcal L^{-1}\left\{F(s)\right\}=e^{−at}f(t)}\]
?????
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