Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.9 The Convolution Integral - Problems - Page 717: 35

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Given: $x(t)=4e^{t}+3\int^t_0 e^{-(t-\tau)}x(\tau)d\tau$ Taking the Laplace transformation: $L[x(t)]=L[4e^{t}+3\int^t_0 e^{-(t-\tau)}x(\tau)d\tau]$ we have: $x(s)=\frac{4}{s-1}+\frac{3}{s+1}L[x(t)]$ Apply convolution theorem: $x(s)=\frac{4s+4}{(s-1)(s-2)}\\ =-\frac{8}{s-1}+\frac{12}{s-22}$ Hence, the general solution is: $x(t)=-8e^{t}+12e^{2t}$