using Laplace transform to solve initial value problems


Since the Laplace transformsMathworldPlanetmath 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 conditionsMathworldPlanetmath 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′′(t)+2f(t)+f(t)=e-t  (f(0)=0,f(0)=1)

gives

[s2F(s)-sf(0)-f(0)]+2[sF(s)-f(0)]+F(s)=1s+1,

i.e.

(s2+2s+1)F(s)=1+1s+1,

whence

F(s)=1(s+1)2+1(s+1)3.

Taking the inverse Laplace transform produces the result

f(t)=te-t+t2e-t2=e-t2(t2+2t).
Title using Laplace transform to solve initial value problems
Canonical name UsingLaplaceTransformToSolveInitialValueProblems
Date of creation 2015-05-29 15:21:45
Last modified on 2015-05-29 15:21:45
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 10
Author pahio (2872)
Entry type Example
Classification msc 34A12
Classification msc 44A10
Related topic TableOfLaplaceTransforms
Related topic LaplaceTransform