Since the Laplace transforms of the derivatives of $f(t)$ are polynomials in the transform parameter $s$ (see table of Laplace transforms), forming the Laplace transform of a linear differential equation with constant coefficients and initial conditions at $t = 0$ yields generally a simple equation for solving the transformed function $F(s)$ . Since the initial conditions can be taken into consideration instantly, one needs not to determine the general solution of the differential equation.

For example, transforming the equation $$f''(t)+2f'(t)+f(t) = e^{-t} \quad (f(0) = 0,\;\; f'(0) = 1)$$ gives $$[s^2F(s)-sf(0)-f'(0)]+2[sF(s)-f(0)]+F(s) = \frac{1}{s+1},$$ i.e. $$(s^2+2s+1)F(s) = 1+\frac{1}{s+1},$$ whence $$F(s) = \frac{1}{(s+1)^2}+\frac{1}{(s+1)^3}.$$ Taking the inverse Laplace transform produces the result $$f(t) \,=\, te^{-t}+\frac{t^2e^{-t}}{2} \;=\; \frac{e^{-t}}{2}(t^2+2t).$$