Question

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

Answer 1

Answer:

Laplace transformation of $L(t^\frac{3}{2}-e^{-10t})$=$\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }\ -\frac{1}{s+10}$

Step-by-step explanation:

Laplace transformation of $L(t^\frac{3}{2}-e^{-10t} )$

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

the integral is now in gamma function form

$\frac{1}{s^{\frac{5}{2}}}\int_{0 }^{\infty}{u} ^\frac{3}{2}e^{-u}{du}=\frac{1}{s^{\frac{5}{2}}}\Gamma(\frac{5}{2})=\frac{1}{s^{\frac{5}{2}}}\times\frac{3}{2}\times\frac{1}{2} }\Gamma (\frac{1}{2} )=\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }$

now laplace of $L(e^{-10t})$

$L(e^{-10t})=\frac{1}{s+10}$

hence

Laplace transformation of $L(t^\frac{3}{2}-e^{-10t})$=$\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }\ -\frac{1}{s+10}$