Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.9 The Convolution Integral - Problems - Page 716: 34

See below

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