Question #1dfad

2 Answers
Feb 14, 2018

#L^(-1) (1/sqrt(2s+3))=1/sqrt(2pix)*e^((-3x)/2)#

Explanation:

#L(s)=1/sqrt(2s+3)#

=#=1/sqrt2*1/sqrt(s+3/2)#

Due to #L[1/sqrt(pix)*e^(-ax)]=1/sqrt(s+a)#

#L^(-1) (1/sqrt(2s+3))=1/sqrt2*1/sqrt(pix)*e^((-3x)/2)#

=#1/sqrt(2pix)*e^((-3x)/2)#

Feb 14, 2018

# ℒ^(-1) \ { 1/sqrt(2s+3) } = e^(-3/2t) /sqrt(2pit) #

Explanation:

We seek:

# ℒ^(-1) \ { 1/sqrt(2s+3) } #

We will need the following standard Laplace transform and inverses:

# {: (ul(f(t)=ℒ^(-1){F(s)}), ul(F(s)=ℒ{f(t)}), ul("Notes")), (af(t), aF(s),a " constant"), (e^(at){1/sqrt(pit)-be^(b^2t)erfc(bsqrt(t))}, 1/(sqrt(s-a)+b),a","b " constant") :} #

So we can write:

# ℒ^(-1) \ { 1/sqrt(2s+3) } = ℒ^(-1) \ { 1/sqrt(2(s+3/2)) } #

# " " = 1/sqrt(2) \ ℒ^(-1) \ { 1/sqrt(s+3/2) } #

# " " = 1/sqrt(2) \ e^(-3/2t){1/sqrt(pit)-0} \ \ \ (a=-3/2,b=0)#

# " " = e^(-3/2t) /sqrt(2pit) #