using Laplace transform to solve initial value problems

Since the Laplace transforms of the derivatives of $f(t)$ arepolynomials in the transform parameter $s$ (see table of Laplacetransforms), forming the Laplace transform of a linear differentialequation with constant coefficients and initial conditions at $t=0$ yields generally a simple equation(image equation (http://planetmath.org/imageequation)) 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

$$f^{\prime \prime }(t)+2f^{\prime }(t)+f(t)=e^{-t}\mathit{\hspace{1em}\hspace{1em}}(f(0)=0,f^{\prime }(0)=1)$$

gives

$$[s^{2}F(s)-sf(0)-f^{\prime }(0)]+2[sF(s)-f(0)]+F(s)=\frac{1}{s+1},$$

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

$$f(t)=t{e}^{-t}+\frac{t^2{e}^{-t}}{2}=\frac{e^{-t}}{2}(t^2+2t).$$