Laplace transform applied to differential equations

Short description: Method for solving linear differential equations using the Laplace transform

In mathematics, the Laplace transform is a powerful integral transform used to switch a function from the time domain to the s-domain. The Laplace transform can be used in some cases to solve linear differential equations with given initial conditions.

## Approach

First consider the following property of the Laplace transform:

ℒ{f′}=sℒ{f}−f(0)

ℒ{f″}=s2ℒ{f}−sf(0)−f′(0)

One can prove by induction that

ℒ{f(n)}=snℒ{f}−∑i=1nsn−if(i−1)(0)

Now we consider the following differential equation:

∑i=0naif(i)(t)=ϕ(t)

with given initial conditions

f(i)(0)=ci

Using the linearity of the Laplace transform it is equivalent to rewrite the equation as

∑i=0naiℒ{f(i)(t)}=ℒ{ϕ(t)}

obtaining

ℒ{f(t)}∑i=0naisi−∑i=1n∑j=1iaisi−jf(j−1)(0)=ℒ{ϕ(t)}

Solving the equation for ℒ{f(t)} and substituting f(i)(0) with ci one obtains

ℒ{f(t)}=ℒ{ϕ(t)}+∑i=1n∑j=1iaisi−jcj−1∑i=0naisi

The solution for f(t) is obtained by applying the inverse Laplace transform to ℒ{f(t)}.

Note that if the initial conditions are all zero, i.e.

f(i)(0)=ci=0∀i∈{0,1,2,... n}

then the formula simplifies to

f(t)=ℒ−1{ℒ{ϕ(t)}∑i=0naisi}

## An example

We want to solve

f″(t)+4f(t)=sin⁡(2t)

with initial conditions f(0) = 0 and f′(0)=0.

We note that

ϕ(t)=sin⁡(2t)

and we get

ℒ{ϕ(t)}=2s2+4

The equation is then equivalent to

s2ℒ{f(t)}−sf(0)−f′(0)+4ℒ{f(t)}=ℒ{ϕ(t)}

We deduce

ℒ{f(t)}=2(s2+4)2

Now we apply the Laplace inverse transform to get

f(t)=18sin⁡(2t)−t4cos⁡(2t)

## Bibliography

* A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1-58488-299-9

0.00

(0 votes)

Original source: https://en.wikipedia.org/wiki/Laplace transform applied to differential equations. Read more

|

Retrieved from "https://handwiki.org/wiki/index.php?title=Laplace_transform_applied_to_differential_equations&oldid=4228610"