Question

Use the convolution theorem to find the inverse Laplace transform of the given function. StartFraction 5 Over s cubed (s squared plus 25 )EndFraction 5 s3s2+25 laplace transform Superscript negative 1 Baseline StartSet StartFraction 5 Over s cubed (s squared plus 25 )EndFraction EndSet (t )ℒ−1 5 s3s2+25(t)

Answer 1

Answer:

$\frac{1}{2}$$t^{2}Sin5t$

Step-by-step explanation:

using the Convolution theorem to find the inverse of :

     $\frac{5}{s^{3}(s^{2}+25 ) }$

   $L^{-1}$  $\frac{5}{s^{3}(s^{2}+25 ) }$ = $\frac{1}{s^{3} }$ × $\frac{5}{s^{2}+25}$

we know from derivation that

Sin(at) =  $\frac{a}{s^{2}+a^{2} }$

Hence: $\frac{5}{s^{2}+25}$  = Sin5t

Also:  $L^{-1}$ $\frac{n!}{s^{n+1} }$ = $t^{n}$

     $L^{-1}$  $\frac{1}{s^{3} }$ = $\frac{1}{2}$ $L^{-1}$ ($\frac{2!}{s^{3} }$)

 = $\frac{1}{2}$$t^{2}$

therefore $L^{-1}$  $\frac{5}{s^{3}(s^{2}+25 ) }$ = $\frac{1}{2}$$t^{2}Sin5t$