Session 4.2 Defining and solving Differential Equations Part b Two ODEs

> restart;

We want to solve two differential equations simultaneously:

dx1/dt = -x1 + ax2 ; x1(0) = 1

dx2/dt = bx1 - 2x2 ; x2(0) = 0

a and b are constants

> de1:=D(x1)(t)=-x1(t)+a*x2(t); The operator notation for differentiation is somewhat more concise than diff.

> de2:=D(x2)(t)=b*x1(t)-2*x2(t);

> ic1:=x1(0)=1;

> ic2:=x2(0)=0;

> s1:=dsolve({de1,de2,ic1,ic2},{x1(t),x2(t)});

The differential equations and initial conditions are all listed in the first set of braces.

The variables to be found are listed in the second set of braces.

We can also solve this equation using a Laplace Transform:

> s2:=dsolve({de1,de2,ic1,ic2},{x1(t),x2(t)},laplace);

Note the addition of the term "laplace" at the end of the command

> assign(s1);x1:=unapply(x1(t),t);x2:=unapply(x2(t),t);

> x1(0);x2(0);

> simplify(x1(0));simplify(x2(0));

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