# 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:

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

[itex]\mathcal{L}\{f''\}
 = s^2 \mathcal{L}\{f\} - s f(0) - f'(0)[itex]

[itex]\mathcal{L}\{f^{(n)}\}
 = s^n \mathcal{L}\{f\} - \Sigma_{i = 1}^{n}s^{n - i}f^{(i - 1)}(0)[itex]


Suppose we want to solve the given differential equation:

[itex]\sum^n_{i=0}a_if^{(i)}(t)=\phi(t)[itex]

This equation is equivalent to

[itex]\sum^n_{i=0}a_i\mathcal{L}\{f^{(i)}(t)\}=\mathcal{L}\{\phi(t)\}[itex]

which is equivalent to

[itex]\mathcal{L}\{f(t)\}={\mathcal{L}\{\phi(t)\}+\sum^n_{i=0}a_i\sum^i_{j=0}s^{i-j}f^{(j-i)}(0) \over \sum^n_{i=0}a_is^i}[itex]

note that the [itex]f^{(k)}(0)[itex] are initial conditions.

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

## An example

We want to solve :

[itex]f^{(2)}(t)+4f(t)=\sin(2t) \,\![itex]

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

we note :

[itex]\phi(t)=\sin(2t) \,\![itex]

and we get :

[itex]\mathcal{L}\{\phi(t)\}=\frac{2}{s^2+4}[itex]

so this is equivalent to :

[itex]s^2\mathcal{L}\{f(t)\}-sf(0)-f^{(1)}(0)+4\mathcal{L}\{f(t)\}=\mathcal{L}\{\phi(t)\}[itex]

we deduce :

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

So we apply the Laplace inverse transform and get

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

## Bibliography

• A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, 2002.

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