# using Laplace transform to solve initial value problems

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 (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

 $f^{\prime\prime}(t)+2f^{\prime}(t)+f(t)=e^{-t}\qquad(f(0)=0,\;\;f^{\prime}(0)=1)$

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

 $f(t)\,=\,te^{-t}+\frac{t^{2}e^{-t}}{2}\;=\;\frac{e^{-t}}{2}(t^{2}+2t).$
Title using Laplace transform to solve initial value problems UsingLaplaceTransformToSolveInitialValueProblems 2015-05-29 15:21:45 2015-05-29 15:21:45 pahio (2872) pahio (2872) 10 pahio (2872) Example msc 34A12 msc 44A10 TableOfLaplaceTransforms LaplaceTransform