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Question

Use the Laplace Method to solve the following IVP:

                                    

Compute     and its Inverse Laplace Transform 

Answer 1

LHS: s²X(s)-sx(0)-x'(0) + 4X(s) = (s²+4)X(s)

RHS: L{tu(t-2)+δ(t-1)} = L{(t-2)u(t-2)+2u(t-2)} + L{δ(t-1)}=[1/s²+2/s]exp(-2s) + exp(-s)

Solving for X(s) we get X(s) = exp(-2s)[1/(s²(s²+4))+2/(s(s²+4))] + exp(-s)/(s²+4).

For the inverse first note that 1/(s²(s²+4)) =1/(4s²) - 1/(4(s²+4)) and 2/(s(s²+4))=1/(2s) - s/(2(s²+4)). Using basic formulas L^-1{exp(-as)L{f}(s)}=u(t-a)f(t-a), L^-1{1/s²}=t, L^-1{1/(s²+4}=sin(2t)/2, L^-1{s/(s²+4}=cos(2t), we get the solution

x(t) = L^-1{X(s)} = u(t - 2)[t/4 - cos(2t - 4)/2 - sin(2t - 4)/8] + u(t - 1)sin(2t - 2)/2

Remark 1: to check that the answer is correct just take the distributional derivatives

Remark 2: if you are good with complex analysis then, instead of partial fraction decomposition, it is simpler just to compute residues.

Answer 2

Per the table, the right side looks like the dirac-delta function with U2(t) and g(t) = t

s^2 Y(t) + 4 Y(t) = exp(-2s) L[t+2]

= exp(-2s) [ 1!/s^2 + 2/s]

Y(t) [ s^2 + 4] = exp(-2s) [ (2s+1)/s^2]

Y(t) = (2s+1)/[ s^2(s^2+4)] (exp(-2s))

partial fraction decomposition results in [ (ax+b)/s^2 + (mx+d)/(s^2+4)] (exp(-2s))

the inverse Laplace gets you U2(t) with driving function sin(2t)