(Root of t - 1/root t)^3

Question

Solve By using laplace transform and its application 

Answer 1

g(t)=(√t-1/√t)³ = t^3/2 -3t^1/2 + 3 /t^-1/2-1/t^-3/2=

=t^3/2-2(t^3/2)'+4(t^3/2)''-(8/3)(t^3/2)'''
If we denote f(t)=t^3/2, then 

g(t)=f(t)-2f'(t)+4f''(t)-(8/3)f'''(t).

Now we can use the addititive and "defivative" properties of Laplace transform:

L{f⁽ⁿ⁾(t)}=sⁿL{f(t)}-s^n-1f(0^-)-s^n-2f'(0^-)-...-f^(n-1)(0^-).

L{g(t)}=L{f(t)}-2L{f'(t)}+4L{f''(t)}-(8/3)L{f'''(t)}=

=L{f(t)}-2[sL{f(t)}-f(0^-)]+4[s²L{f(t)}-sf'(0^-)-f(0^-)]-

-(8/3)[s³L{f(t)}-s²f''(0^-) -sf'(0^-)-f(0^-)]=

=(1-2s+4s²-(8/2)s³)L{f(t)}+[2f(0^-)-4sf'(0^-)-4f(0^-)+(8/3)s²f'''(0^-)-(8/3)sf''(0^-)+(8/3)f(0^-)]= (**)

 

L{f(t)}=L{t^3/2}=∫₀^∞e^-stt^3/2dt=(3/4)√π * s^-5/2.

f(0^-)=f'(0^-)=0, but f''(0^-)=∞.

 

Now plug in all values in (**) and simplify. The rest I leave for you to do.