# 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

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

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

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

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 |