Question

Use the transforms in section 4.1 to find the Laplace transform of the function. t^3/2 - e^-10t

Answer 1

Laplace transformation of
`L(t^(3)/(2)-e^(-10t))`=
`(3√(\pi) )/(4 s^(5)/(2) )\ -(1)/(s+10)`

Explanation:

Laplace transformation of
`L(t^(3)/(2)-e^(-10t)  )`

`L(t^(3)/(2))=\int_(0 )^(\infty)t^(3)/(2)e^(-st)dt\\substitute \ u =st\\L(t^(3)/(2))=\int_(0 )^(\infty)(u)/(s) ^(3)/(2)e^(-u)(du)/(s)=\frac{1}{s^{(5)/(2)}}\int_(0 )^(\infty){u} ^(3)/(2)e^(-u){du}`

the integral is now in gamma function form

`\frac{1}{s^{(5)/(2)}}\int_(0 )^(\infty){u} ^(3)/(2)e^(-u){du}=\frac{1}{s^{(5)/(2)}}\Gamma((5)/(2))=\frac{1}{s^{(5)/(2)}}*(3)/(2)*(1)/(2) }\Gamma ((1)/(2) )=(3√(\pi) )/(4 s^(5)/(2) )`

now laplace of
`L(e^(-10t))`

`L(e^(-10t))=(1)/(s+10)`

hence

Laplace transformation of
`L(t^(3)/(2)-e^(-10t))`=
`(3√(\pi) )/(4 s^(5)/(2) )\ -(1)/(s+10)`