Laplace transforms and the solution to linear ordinary differential equations

<< Calculus`LaplaceTransform`

Laplace Transforms of some elementary functions,
and linearity.

Some elementary functions

\(\(
LaplaceTransform[Exp[a\ t], \ t, \ s]\)\)

\(LaplaceTransform[Sin[a\ t], \ t, \ s]
\)LaplaceTransform[Cos[a t],  t, s]

\(
LaplaceTransform[\ Exp[\ a\ t]\ Sin[b\ t], \ t, \ s]\)LaplaceTransform[ Exp[ a t] Cos[b t],  t,  s]

\(
LaplaceTransform[\ t^3\ Exp[a\ t], \ t, s]
\)\(LaplaceTransform[\ t^4\ \ Exp[a\ t], \ t, s]
\)

\(
LaplaceTransform[Sinh[a\ t], \ t, \ s]\)LaplaceTransform[Cosh[a t],  t,  s]

Powers of    t   (and the Gamma function).

LaplaceTransform[t^7,  t, s]Factorial[7]

LaplaceTransform[t^(3.34),  t, s]Gamma[4.34]

LaplaceTransform[t^(-1/2),  t, s]

Linearity

LaplaceTransform[c1 f[t]  +  c2 g[t],  t,  s]

The Laplace transform and derivatives.

LaplaceTransform[Derivative[1][f][t],  t, s]

LaplaceTransform[Derivative[2][f][t],  t, s]

LaplaceTransform[Derivative[3][f][t],  t, s]

Examples the use of the Laplace transform to solve  initial value problems.

Consider the initial value problem:
y'' + y = Sin[2t]
y(0) = 2;   y'(0) = 1.
(Cf. Example 1, page 301 of Boyce and Diprima.)

Clear[F, f, y, L,  LT]

\(\(
L[y_]\  = \ \(\(Derivative[2]\)[y]\)[t]\  + \ y[t]
\)\)

Call the solution to the initial value problem  f[t], and call its Laplace transform F[s].  We compute the Laplace transform of  L[f] - Sin[2t]  in terms of the Laplace transform of    f,    then substitute in the intial values f[0] = 2,  f'[0] = 1

L[f]  -  Sin[2  t]\(f[0]\  = \ 2; 
\(f'\)[0]\  = \ 1; 


    LT\  = \ LaplaceTransform[L[f]\  - \ Sin[2  t], t, s]\  //. 

        		{LaplaceTransform[f[t], \ t, s]\  -> \ F}

		\)

We set this equal to zero and solve for F.

Solve[LT == 0,  F]F  =  Simplify[ F //.  % [[1, 1]]]

The solution to the initial value problem is the inverse Laplace transform of F.

\(
F\  = \ Expand[Apart[F]]\)InverseLaplaceTransform[F,  s, t]

Let's check that we have found the solution.

Clear[h]h[t_]  =  InverseLaplaceTransform[F,  s, t]L[h]

Consider the initial value problem:
Derivative[4][y]-y=0
y(0) = 0;   y'(0) = 1; y''(0) = 0;  y'''(0) = 0.
(Cf. Example 2, page 301 of Boyce and Diprima.)

Clear[F, f, y, L,  LT]L[y_]  =  Derivative[4][y][t]  -  y[t]\(	f[0]\  = \ 0; 
\(f'\)[0]\  = \ 1; 
\(\(f'\)'\)[0]\  = \ 0; 

    \(\(\(f'\)'\)'\)[0]\  = \ 0; 

    LT\  = \ LaplaceTransform[L[f], t, s] //. \ 
						
        LaplaceTransform[f[t], \ t, s]\  -> \ F\)Solve[LT  ==  0,  F]\(F\  = \ \ F //. %\ [\([1, 1]\)]
\)

The solution to the initial value problem is the inverse Laplace transform of F.

\(
\ Expand[Apart[F]]\)InverseLaplaceTransform[F,  s, t]