San José State University
Math 133A, Fall 2005
Solutions to graded Homework 3 problems
Ex. 1.5, #8. Let f (y) = (y − 2)(y − 3)y. Then our ODE is dy/dt = f (y). The equilibria are 0, 2,
and 3, since f equals zero there. Does the (unique) solution y(t) such that y(0) = −1 increase
or decrease? Since f (−1) = (−3)(−4)(−1) = −12 < 0, y(t) decreases as t increases. Therefore,
y(t) → −∞, as t → ∞. Otherwise, y(t) would have to converge to some y∗ < −1, which is
impossible, since y∗ would have to be an equilibrium (and there are no equilibria less than 0).
Ex. 1.5, #12. (a) We have, for i = 1, 2:
1
dyi
=−
dt
(t − i)2
= −yi (t)2 ,
so y1 and y2 satisfy the ODE.
(b) Since y1 (0) = −1 and y2 (0) = −1/2, at t = 0, y(t) is between y1 (t) and y2 (t) when t = 0. By
the Uniqueness Theorem, y(t) cannot cross any other solutions, so y(t) must remain between
y1 (t) and y2 (t) for all t for which y(t), y1 (t), and y2 (t) are defined. In other words, for all such
t, we have
y1 (t) < y(t) < y2 (t).
(Note. It’s not too hard to see that this holds for t < 1.)
Ex. 1.6, #12. The equilibria are w = −1, 0, 1, since arctan w = 0 if and only if w = 0. Let
f (w) = (w2 − 1) arctan w. Then f (w) > 0 for w > 1 and −1 < w < 0; f (w) < 0 for 0 < w < 1
and w < −1. (You can check this by simply plugging in a “test value” for w into f (·)).
Therefore, the phase line looks like this:
1
0
−1
It follows that −1 and 1 are sources, and 0 is a sink.
1
Ex. 1.6, #19. The quadratic polynomial w2 + 2w + 10 has no real roots (its roots are −1 ± 3i), so
it is either always positive or always negative. It’s not hard to see that it must be the former
(plug in w = 0 and get 10 > 0). Therefore, all solutions to the ODE dw/dt = w2 + 2w + 10
are always increasing and approach infinity, as t → ∞. The solutions satisfying w(0) = 0,
w(1/2) = 1, and w(0) = 2 are pictured in the following figure:
NOT GRADED:
p
Ex. 1.5, #10. (a) If we plug the function y(t) ≡ 0 into the ODE dy/dt = 2 |y|, we see that it
satisfies it.
(b) Assume first that y > 0. Separating the variables, we obtain y = (t−C)2 , for some constant
C. Checking whether this is a solution, we see that it is, but only for t > C. (This is why the
absolute value sign in an ODE can be dangerous.) If y < 0, then |y| = −y. Again by separation
of variables, we get y = −(t + C)2 , for some constant C. Plugging into the equation, we see
that this a solution only for t < C.
Therefore, for any real number C, we get these solutions: y0 (t) ≡ 0,
(
(
−(t − C)2
0
if t < C
y
(t)
=
y1 (t) =
2
0
(t − C)2 if t ≥ C,
and
y3 (t)
(
−(t − C)2
(t − C)2
if t < C
if t ≥ C,
if t < C
if t ≥ C.
(c) The reason
p this doesn’t contradict the Uniqueness Theorem is that the right-hand side of
the ODE,p2 |y| does not satisfy the hypotheses of the Uniqueness Theorem. Namely, the
function |y| does not have a derivative at y = 0.
2
(d) Given the initial condition y(0) = 0 (or y(t0 ) = 0), HPGSolver plots the solution y0 . For
any initial condition y(0) 6= 0, it plots y3 . This can be see from the following plot:
Ex. 1.5, #14. (a) Separating the variables, we obtain y(t) = ±(C − 2t)−1/2 . Using the initial
condition y(0) = 1, we get C = 1, so the solution is
1
y(t) = + √
.
1 − 2t
(b) The solution is defined only when 1 − 2t > 0, i.e., for t < 1/2.
√
(c) As t → 1/2 from the left, 1 − 2t approaches 0 from the right (i.e., through positive values),
so y(t) → +∞. Since it “blows up”, there is no way y(t) can be extended in a meaningful way
for more time. Note, however, that if its limit was finite, as t approaches the boundary of the
domain of definition, then we could extend y(t) continuously for more time.
3