Why does the laplace transform not work this way

Question

If there is a differential equation $$(D^2+5D+6)y(t) = (D+1) x(t) $$ where $$x(t) = e^{-4t} $$ In order to solve the differential equation why is it that the Laplace transform will be $$S^2Y(S)-SY(0^-)-Y'(0^-) +5SY(S)-5Y(0^-)+6Y(S) =\frac{S}{S+4}+\frac{1}{S+4} $$ Why can we not derivate x(t) in the 't' domain and then find its Laplace transform to solve the differential equation? i.e. why is the below form incorrect $$S^2Y(S)-SY(0^-)-Y'(0^-) +5SY(S)-5Y(0^-)+6Y(S) =\frac{-3}{S+4}$$ since $$(D+1)x(t) = -4e^{-4t}+e^{-4t} = -3e^{-4t}$$

Answer 1

The Laplace transform of

$$D[e^{-4t}]$$

is $$S\mathcal{L}(e^{-4t}) - e^{-4\times0} = \frac{S}{S+4} - 1 = \frac{-4}{S+4}.$$

If you make this correction it agrees with your second calculation.