Question

In Problems 3 - 24 fill in the blanks or answer true or false.

17. ${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{s}{s^2-10s+29}\right\}$=……

Short answer

$L^{-1}\left\{\frac{s}{s^2-10s+29}\right\}=e^{5t}\cos \left(2t\right)+\frac{5}{2}{e}^{5t}\sin \left(2t\right)$

Step-by-step solution

Consider the function

The function is,

$L^{-1}\left\{\frac{s}{s^2-10s+29}\right\}$

Separate the function,

$\frac{s}{s^2-10s+29}=\frac{s-5}{{\left(s-5\right)}^2+4}+\frac{5}{{\left(s-5\right)}^2+4}$

thus, $L^{-1}\left\{\frac{s}{s^2-10s+29}\right\}=L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}+\frac{5}{{\left(s-5\right)}^2+4}\right\}$

Compute the Laplace inverse

Consider,

$L^{-1}\left\{\frac{s}{s^2-10s+29}\right\}=e^{5t}\cos \left(2t\right)+\frac{5}{2}{e}^{5t}\sin \left(2t\right)$

Compute the Laplace inverse of the first term

Consider

$L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}\right\}$

$\mathrm{If} L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right) \mathrm{then} L^{-1}\left\{F\left(s-a\right)\right\}=e^{at}f\left(t\right)$

for$\frac{s-5}{{\left(s-5\right)}^2+4}:a=5, F\left(s\right)=\frac{s}{s^2+4}$

Thus,$L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}\right\}=e^{5t}{L}^{-1}\left\{\frac{s}{s^2+4}\right\}$

here,$L^{-1}\left\{\frac{s}{s^2+4}\right\}=\cos \left(2t\right)$

Thus,$L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}\right\}=e^{5t}\cos \left(2t\right)$

Compute the Laplace inverse of the second term

Consider,

$5L^{-1}\left\{\frac{1}{{\left(s-5\right)}^2+4}\right\}$

$\mathrm{If} L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right) \mathrm{then} L^{-1}\left\{F\left(s-a\right)\right\}=e^{at}f\left(t\right)$

for,$\frac{5}{{\left(s-5\right)}^2+4}:a=5, F\left(s\right)=\frac{5}{s^2+4}$

Thus,$L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}\right\}=e^{5t}{L}^{-1}\left\{\frac{5}{s^2+4}\right\}$

Here,$L^{-1}\left\{\frac{5}{s^2+4}\right\}=\frac{5}{2}\sin \left(2t\right)$

Thus,$L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}\right\}=e^{5t}\frac{5}{2}\sin \left(2t\right)$

Compute the Laplace inverse

Consider the function,

$L^{-1}\left\{\frac{s}{s^2-10s+29}\right\}=L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}+\frac{5}{{\left(s-5\right)}^2+4}\right\}$

$\Rightarrow L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}+\frac{5}{{\left(s-5\right)}^2+4}\right\}=L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}\right\}+5L^{-1}\left\{\frac{1}{{\left(s-5\right)}^2+4}\right\}$

Therefore, the Laplace inverse of$L^1\left\{\frac{s}{S^2-10s+29}\right\}$

$\therefore L^{-1}\left\{\frac{s-5}{{\left(s-5\right)}^2+4}\right\}+5L^{-1}\left\{\frac{1}{{\left(s-5\right)}^2+4}\right\}=e^{5t}\cos \left(2t\right)+e^{5t}\frac{5}{2}\sin \left(2t\right)$