Linear differential equations systems with constant coefficients.

  Example: suppose we want to solve the following system:

with initial data y₁(0)=y₂(0)=0.

For scalar linear differential equations with constant coefficients, it is a well-known procedure to use Laplace transform, e.g. to solve

one would perform the L-transform and get:

where f(0) and f'(0) are the initial consitions at x=0. If L(f) is a rational fraction, INT ILAP allows you to recover y by inverse Laplace transform of

But this method may be applied to systems of linear differential equations, the aim of LDEC is to help you solving 1st order systems (note that higher order systems may always be rewritten as 1st order systems). Let y=(y₁, ..., y_n) be a vector of functions of x and suppose that we want to solve:

where A is a constant matrix and b a vector of n functions of x Denote by L(b) the vector of the n Laplace transform of the n functions of b , and by y(0) the vector of n initial data at x=0 . Then

and:

The purpose of LDEC is to compute

given A and L(b)+y(0).
stk2: A,
stk1: L(b)+y(0) LDEC -> ( p Id - A )^-1 ( L(b) + y(0) )

Back to the example above: The matrix A is then: [ [ 1 -1 ] [ 2 4 ] ] and since Laplace(1)=1/p and Laplace(e^x)=1/(p-1), we put L(b)+y(0) on the stack: { '1/X' '1/(X-1)' } (here VX is set as usual to 'X') After calling LDEC, we get:
{ '(X^2-6*X+4)/(X^2-X)/((X-3)*(X-2))' '(X+2)/X/((X-3)*(X-2))' } which is the { Laplace(y₁ ) Laplace(y₂) }. Now to recover y₁ or/and y₂ just hit EVAL and INT ILAP for each coordinate. You get:

To compute the solution of the LDEC with initial data y₁(0)=1, y₂(0)=2 just replace { '1/X' '1/(X-1)' } by { '1/X+1' '1/(X-1)+2' }..