David U. answered • 08/20/20

Scientist, Math PhD and Physics PhD, tutoring students/professionals

LHS: *s*^{2}*X(s)-sx(0)-x'(0) + 4X(s) = (s*^{2}*+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/s*^{2}*+2/s]exp(-2s) + exp(-s)*

Solving for *X(s)* we get *X(s) = exp(-2s)[1/(s*^{2}*(s*^{2}*+4))+2/(s(s*^{2}*+4))] + exp(-s)/(s*^{2}*+4).*

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