Laplace transform applied to differential equations

The use of Laplace transform makes it much easier to solve linear differential equations with given initial conditions.

First consider the following relations:: $mathcal\{L\}\{f\text{'}\}\; =\; s\; mathcal\{L\}\{f\}\; -\; f(0)$: $mathcal\{L\}\{f"\}\; =\; s^2\; mathcal\{L\}\{f\}\; -\; s\; f(0)\; -\; f\text{'}(0)$: $mathcal\{L\}\{f^\{(n)\}\}\; =\; s^n\; mathcal\{L\}\{f\}\; -\; Sigma\_\{i\; =\; 1\}^\{n\}s^\{n\; -\; i\}f^\{(i\; -\; 1)\}(0)$

Suppose we want to solve the given differential equation:: $sum^n\_\{i=0\}a\_if^\{(i)\}(t)=phi(t)$

This equation is equivalent to

: $sum^n\_\{i=0\}a\_imathcal\{L\}\{f^\{(i)\}(t)\}=mathcal\{L\}\{phi(t)\}$

which is equivalent to

$mathcal\{L\}\{f(t)\}=\{mathcal\{L\}\{phi(t)\}+sum^n\_\{i=0\}a\_isum^i\_\{j=1\}s^\{i-j\}f^\{(j-1)\}(0)\; over\; sum^n\_\{i=0\}a\_is^i\}$

note that the $f^\{(k)\}(0)$ are initial conditions.

Then all we need to get "f"("t") is to apply the Laplace inverse transform to $mathcal\{L\}\{f(t)\}$

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

: $phi(t)=sin(2t)\; ,!$

and we get

: $mathcal\{L\}\{phi(t)\}=frac\{2\}\{s^2+4\}$

So this is equivalent to

: $s^2mathcal\{L\}\{f(t)\}-sf(0)-f^\{\; \text{'}\}(0)+4mathcal\{L\}\{f(t)\}=mathcal\{L\}\{phi(t)\}$

We deduce: $mathcal\{L\}\{f(t)\}=frac\{2\}\{(s^2+4)^2\}$

So we apply the Laplace inverse transform and get

:$f(t)=frac\{1\}\{8\}sin(2t)-frac\{t\}\{4\}cos(2t)$

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

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