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:

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

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

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

Suppose we want to solve the given differential equation:

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

This equation is equivalent to

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

which is equivalent to

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

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

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

We deduce

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

So we apply the Laplace inverse transform and get

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

order of derivative is the power of "S" with first term and then decrease power of "S". I f order is two total terms in expansion will be three and so on.

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

External links

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