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12.4.2  Laplace transform

Denoting by L the Laplace transform, you get the following:

     
  L(y)(x)
=
+∞


0
ex uy(u)du 
         
L−1(g)(x)
=
1
2iπ
 


C
 ez xg(zdz
         

where C is a closed contour enclosing the poles of g.

The laplace command finds the Laplace transform of a function.

The ilaplace or invlaplace command finds the inverse Laplace transform of a function.

You also can use the addtable command Laplacians of unspecified functions (see Section 19.4.3). The Laplace transform has the following properties:

     
  L(y′)(x)=−y(0)+xL(y)(x)         
L(y′′)(x)=−y′(0)+xL(y′)(x)         
 =−y′(0)−xy(0)+x2L(y)(x)          

These properties make the Laplace transform and inverse Laplace transform useful for solving linear differential equations with constant coefficients. For example, suppose you have

     
 y′′ +p y′ +q y=f(x)         
 y(0)=a,  y′(0)=b          

then

     
  L(f)(x)=L(y′′+py′+qy)(x         
 =−y′(0)−x y(0)+x2 L(y)(x)−p y(0)+p x L(y)(x))+q L(y)(x         
 =(x2+p x+qL(y)(x)−y′(0)−(x+py(0)          

Therefore, if a=y(0) and b=y′(0), you get

  L(f)(x)=(x2+p x+q)L(y)(x)−(x+pab

and the solution of the differential equation is:

  y(x)=L−1


L(f)(x)+(x+pa +b
x2+p x+q



.

Examples

laplace(sin(x))
     
1
x2+1
          

With t as the original variable:

laplace(sin(t),t)
     
1
t2+1
          

With t as the original variable and s as the transform variable:

L:=laplace(sin(t),t,s)
     
1
s2+1
          

Apply the inverse transform to obtain the original function:

ilaplace(L,s,t)
     
sint           

Solve y′′ −6 y′+9 y=x e3 x, y(0)=c0, y′(0)=c1. Here, p=−6 and q=9.

laplace(x*exp(3*x))
     
1
x2−6 x+9
          
ilaplace((1/(x^2-6*x+9)+(x-6)*c_0+c_1)/(x^2-6*x+9))
     
1
6
 
x3−18 x c0+6 x c1+6 c0
ex
          

Note that this equation could be solved directly.

desolve(y''-6*y'+9*y=x*exp(3*x),y)
     
ex 
c0 x+c1
+
1
6
 x3 ex
          

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