Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.9 The Convolution Integral - Problems - Page 717: 36

See below

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