MTH 264
DELTA COLLEGE
SECTION 1.5 9
(a) Show that y1 (t) = t2 and y2 (t) = t2 + 1 are solutions to the differential equation.
dy
= −y 2 + y + 2yt2 + 2t − t2 − t4
dt
(b) Show that if y(t) is also a solution to the differential equation in part (a) and if 0 < y(0) < 1, then
t2 < y(t) < t2 + 1 for all t.
(c) Illustrate your answer using HPGSolver.
Solution:
(a)
d
(y1 (t)) = 2t
dt
and
−(y1 (t))2 + (y1 (t)) + 2(y1 (t))t2 + 2t − t2 − t4
=
−(t2 )2 + (t2 ) + 2(t2 )t2 + 2t − t2 − t4
=
−t4 + t2 + 2t4 + 2t − t2 − t4
=
2t
d
(y2 (t)) = 2t
dt
and
−(y2 (t))2 + (y2 (t)) + 2(y2 (t))t2 + 2t − t2 − t4
= −(t2 + 1)2 + (t2 + 1) + 2(t2 + 1)t2 + 2t − t2 − t4
= −t4 − 2t2 − 1 + t2 + 1 + 2t4 + 2t2 + 2t − t2 − t4
=
2t
So we have shown directly that both y1 (t) and y2 (t) are solutions of the given differential equation.
(b) Notice that this differential equation meets the conditions of the existence and uniqueness theorems; even
that
f (t, y) = −y 2 + y + 2yt2 + 2t − t2 − t4
and
∂f
= −2y + 1 + 2t2
∂y
are continuous everywhere in the (t, y)-plane.
Next observe that y1 (t) and y2 (t) are solutions to the differential equations with initial conditions of y1 (0) = 0
and y2 (0) = 1, respectively and that y1 (t) and y2 (t) are both solutions over the whole real line −∞ < t < ∞.
Now suppose that we seek a solution of the given
equation with an initial value y(0) = c, 0 < c < 1,
represented by the red circle on the y-axis in the figure
on the right. We can know with certainty that whatever
solution passes through that point will stay bound by the
two blue solution curves that we have verified above in
(a) for as long as it is defined. Why? Such a solution will
exist over a fixed interval − < t < because we have met
the conditions of the existence theorem, and if the solution
y(t) shared a value at any time in this interval with one of
the solutions y1 (t) or y2 (t), say y(a) = y2 (a), − < a < ,
then according to the uniqueness theorem, y(t) and y2 (t)
would have to agree on a whole interval, whatever interval
is common between − < t < and the interval −∞ < t < ∞ where y2 exists as a solution. In other words
y(t) and y2 (t) would have to agree wherever they are both defined , and since they are both defined at t = 0 they
would have to agree there as well.
That is a contradiction however, because at t = 0, y(0) = c < 1 = y2 (0).
Conclusion: if another solution starts with an initial condition y(0) = c < 1, then that solution must remain
below y2 (t) as long as it is defined. Likewise, if 0 < c, then that solution must remain above y1 (t) as long as it
is defined. It cannot cross either of these two “boundary” curves.
So when we find this new solution, and we know that we can, it will remain between y1 (t) and y2 (t) as long
as it is defined, that is
y1 (t) = t2 < y(t) < t2 + 1 = y2 (t)
for all t for which y(t) is defined. Below is an image from the HPGSolver: