87 Example, concluded

which we write as partial fractions

$$\widehat{y}(s)=\frac{A}{s-a}+\frac{B}{s-b-1};$$

then

$$A=\frac{2(a-b)}{a-b-1};B=\frac{-2}{a-b-1};$$

by the uniqueness of Laplace transforms, we have a unique solution

$$y(t)=\frac{2(a-b)e^{at}-2e^{(b+1)t}}{a-b-1}.$$