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:

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

One can prove by induction that

L { f ( n ) } = s n L { f } − ∑ i = 1 n s n − i f ( i − 1 ) ( 0 ) {\displaystyle {\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:

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

with given initial conditions

f ( i ) ( 0 ) = c i {\displaystyle f^{(i)}(0)=c_{i}}

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

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

obtaining

L { f ( t ) } ∑ i = 0 n a i s i − ∑ i = 1 n ∑ j = 1 i a i s i − j f ( j − 1 ) ( 0 ) = L { ϕ ( t ) } {\displaystyle {\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 L { f ( t ) } {\displaystyle {\mathcal {L}}\{f(t)\}} and substituting f ( i ) ( 0 ) {\displaystyle f^{(i)}(0)} with c i {\displaystyle c_{i}} one obtains

L { f ( t ) } = L { ϕ ( t ) } + ∑ i = 1 n ∑ j = 1 i a i s i − j c j − 1 ∑ i = 0 n a i s i {\displaystyle {\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 L { f ( t ) } . {\displaystyle {\mathcal {L}}\{f(t)\}.}

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

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

then the formula simplifies to

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