Laplace Transform to ODEs

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.

First consider the following property of the Laplace transform:

{\mathcal {L}}\{f'\}=s{\mathcal {L}}\{f\}-f(0)

{\mathcal {L}}\{f''\}=s^{2}{\mathcal {L}}\{f\}-sf(0)-f'(0)

One can prove by induction that

{\mathcal {L}}\{f^{(n)}\}=s^{n}{\mathcal {L}}\{f\}-\sum _{i=1}^{n}s^{n-i}f^{(i-1)}(0)

Now we consider the following differential equation:

\sum _{i=0}^{n}a_{i}f^{(i)}(t)=\phi (t)

with given initial conditions

f^{(i)}(0)=c_{i}

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

\sum _{i=0}^{n}a_{i}{\mathcal {L}}\{f^{(i)}(t)\}={\mathcal {L}}\{\phi (t)\}

obtaining

{\mathcal {L}}\{f(t)\}\sum _{i=0}^{n}a_{i}s^{i}-\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}f^{(j-1)}(0)={\mathcal {L}}\{\phi (t)\}

Solving the equation for ${\mathcal {L}}\{f(t)\}$ and substituting $f^{(i)}(0)$ with $c_{i}$ one obtains

{\mathcal {L}}\{f(t)\}={\frac {{\mathcal {L}}\{\phi (t)\}+\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}c_{j-1}}{\sum _{i=0}^{n}a_{i}s^{i}}}

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

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

f^{(i)}(0)=c_{i}=0\quad \forall i\in \{0,1,2,...\ n\}

then the formula simplifies to

f(t)={\mathcal {L}}^{-1}\left\{{{\mathcal {L}}\{\phi (t)\} \over \sum _{i=0}^{n}a_{i}s^{i}}\right\}

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}}

The equation is then equivalent to

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

We deduce

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

Now we apply the Laplace inverse transform to get

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

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