3.1 Homogeneous Equations with Constant
Coefficients
Maple Setup
In this session, we will use solve, dsolve, and plot. The command assign is a
convenient way to give values to variables.
Problem 25
25. Consider the initial value problem
2y'' + 3y' -2y = 0,
y(0) = 1,
y'(0) = -
where > 0.
(a) Solve the initial value problem.
(b) Plot the solution when
. Find the coordinates (
) of the minimum point
of th solution in this case.
(c) Find the smallest value of for which teh solution has no minimum point.
Begin by defining the initial value problem ivt. Use y0 and yp0 for the initial values.
> de := 2*diff(y(t),t,t)+3*diff(y(t),t)-2*y(t) = 0;
y0 := 1;
yp0 := -beta;
The characterstic equation ce is
> ce := 2*r^2+3*r-2=0;
We know that solutions of de must be y =
> r := 'r';
r := [solve(ce, r)];
.
.
soln := t -> c[1]*exp(r[1]*t)+c[2]*exp(r[2]*t);
Substituting the initial values determines
and
.
> sys := {soln(0)=y0, D(soln)(0)=yp0};
the_constants := solve(sys, {c[1],c[2]});
assign(the_constants);
The solution is
> soln(t);
Of course, we could have used Maple's dsolve. Do it to verify our computation.
> dsolve({de, y(0)=y0, D(y)(0)=yp0}, y(t));
Set
and plot soln.
> beta := 1;
plot(soln(t), t=-1..3, 0..5);
From elementary calculus, we know that the minimum point is where
.
> Min := solve(diff(soln(t), t)=0, t):
MinPoint := [Min, soln(Min)];
evalf(MinPoint);
Clear 's value and consider soln(t).
> beta := 'beta';
soln(t);
If
> 2, then the coefficient of the exponential growth term
become large with large t and must have a minimum. If
is positive; soln will
, this term vanishes and
there is only exponential decay with no minimum, but limit 0. If < 2, then the growth
term subtracts and also prvents a minimum. Try several values to convince yourself.