# Reverse Laplace transform of \frac{3s-15}{2s^{2}-4s+10}

Reverse Laplace transform of
$\frac{3s-15}{2{s}^{2}-4s+10}$
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Fasaniu
Step 1
$\frac{3s-15}{2{s}^{2}-4s+10}=\frac{3}{2}\frac{s-5}{{s}^{2}-2s+5}=\frac{3}{2}\frac{s-1-4}{{\left(s-1\right)}^{2}+4}$
$=\frac{3}{2}\frac{s-1}{{\left(s-1\right)}^{2}+{2}^{2}}-3\frac{2}{{\left(s-1\right)}^{2}+{2}^{2}}$
Now let .
The inverse transforms of , respectively. Hence the inverse transform of .

lovagwb
First of all, get rid of that 2 factor in the denominator: $2{s}^{2}-4s+10=2\left({s}^{2}-2s+5\right)$ and observe that the polynomial ${s}^{2}-2s+5$ has no real roots. So, we write it as follows:
${s}^{2}-2s+5={\left(s-1\right)}^{2}+\left(5-1\right)={\left(s-1\right)}^{2}+{2}^{2}$
Next, we try to decompose our fraction like this:
$\frac{3s-15}{{\left(s-1\right)}^{2}+{2}^{2}}=A\frac{s-1}{{\left(s-1\right)}^{2}+{2}^{2}}+B\frac{2}{{\left(s-1\right)}^{2}+{2}^{2}}$,
for some constants A and B which youll

alenahelenash
Pulling out a factor of 2 from the denominator and the constant multiple of 3 from the numerator leads to: $\frac{3s-15}{2{s}^{2}-4s+10}=\frac{3}{2}\frac{s-5}{{s}^{2}-2s+5}$ Next we do Completing the Square for the polynomial $\left({s}^{2}-2s+5\right)$ in the denominator and re-writing the numerator constant giving us: $=\frac{3}{2}\cdot \frac{s-1-4}{\left(s-1{\right)}^{2}+4}$ Let denominator equal K. Separating the fraction into parts with them being ; after distributing the $\frac{3}{2}$ to the second term −4. This leads to the expression below now. $=\frac{3}{2}\cdot \frac{\left(s-1\right)}{\left(s-1{\right)}^{2}+{2}^{2}}-3\cdot \frac{2}{\left(s-1{\right)}^{2}+{2}^{2}}$ From the table of transforms, the inverse transform of are , respectively. Hence the inverse laplace transform of our original problem is, ${L}^{-1}\left\{\frac{3-15}{2{s}^{2}-4s+10}\right\}=\frac{3}{2}{e}^{t}\mathrm{cos}\left(2t\right)-3{e}^{t}\mathrm{sin}\left(2t\right)$