Differential EquationsHardLaplace Transforms ShareLaplace Transforms · Hard problemUsing L{tf(t)}=−F′(s)\mathcal{L}\{t f(t)\} = -F'(s)L{tf(t)}=−F′(s), find L{tsin2t}\mathcal{L}\{t\sin 2t\}L{tsin2t} evaluated at s=1s = 1s=1.Check Show answer Walk through it step by step Practice more like this Review this concept