Math 5410 Exam 2 Solutions
1) Notice that et cos 2t = Re[e(1+2i)t ]. Therefore, we can find a real-valued
solution yd by finding a complex-valued solution zd of
z 00 + 2z 0 + 4z = e(1+2i)t
and then letting yd = Re[zd ]. We can look for a solution of the form zd (t) =
Ce(1+2i)t for some constant C. Plugging zd into the equation and canceling
the complex exponential from both sides, we obtain that (3 + 8i)C = 1, or
C=
Therefore, zd (t) =
3−8i (1+2i)t
e
,
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yd (t) =
1
3 − 8i
=
3 + 8i
73
so that taking the real part, we get
3 t
8
e cos 2t + et sin 2t.
73
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The characteristic equation
for this operator is r 2 + 2r + 4 = 0. This
√
equation has roots −1 ± i 3. Therefore the general solution of the undriven
equation is
√
√
C1 e−t cos 3t + C2 e−t sin 3t
where C1 and C2 are arbitrary constants. Since the general solution for the
driven equation is the sum of yd and the general solution of the undriven
equation, we see that there is no choice of C1 and C2 for which a solution
of the driven equation is periodic. Regardless of the choice of C1 and C2 ,
the presence of the negative exponential (and the boundedness of the trig
functions) makes the contribution of the solution to the undriven equation
negligible for large t, so for large t, all solutions look like yd , which is not
periodic because its oscillations increase without bound.
Another way to see that there can be no periodic solution is to note that
if y is periodic, then so are y 0 and y 00 . Therefore, so is the left-hand side
of our equation, but the right hand-side is not, so there can be no periodic
solutions.
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2a) The set S = {y ∈ C 2 (R ) : L[y] = f } where L is the operator given by
L[y] = y 00 + a(t)y 0 + b(t)y,
cannot be empty because given any t0 , y0 and y00 in R , the IVP L[y] = f ,
y(t0 ) = y0 , y 0(t0 ) = y00 has a unique solution (the hypotheses of the Existence
and Uniqueness Theorem hold), which must be in S. In fact, the set S has
infinitely many elements.
S is not a vector space. Let y1 and y2 be in S. Because L is linear,
L[y1 + y2 ] = L[y1 ] + L[y2 ] = f + f = 2f.
For y1 +y2 to be in S, we must have L[y1 +y2 ] = f . Because f is nonconstant,
it is not identically zero, so f 6= 2f . Therefore L[y1 + y2 ] = 2f 6= f and
y1 + y2 6∈ S, and S is not a vector space.
2b) Let y(t) be the unique solution to the IVP y 0 + p(t)y = f , y(0) = 1.
This problem has a solution by the existence and uniqueness theorem for
linear first-order equations. The theorem applies because fn → f uniformly
on [0, ∞), so f is continuous. The initial condition was chosen because if yn
is going to converge to y uniformly, it will also converge pointwise at each
point, in particular at t = 0. We are given that yn (0) = 1 for all n, so y(0)
has to be 1. Now let zn = yn − y. Then:
zn0 = yn0 − y 0 = −p(t)(yn − y) + fn − f
Also, zn (0) = 0 for all n. In other words, zn solves the IVP zn0 + p(t)zn =
fn − f , zn = 0. Then by Theorem 2.3.1 (BIBO), with p0 = 1 and M = 1/n,
we get that
1
|zn (t)| = |yn (t) − y(t)| ≤
n
for all t ≥ 0, which implies that yn → y uniformly.
3a) The only equilibrium solution is y ≡ 0. Any equilibrium must satisfy
ey − 1 = 0 since the first term in the right-hand side of the equation will be
0 for any constant function y. The only solution of ey − 1 = 0 is y ≡ 0.
3c) The only equilibrium point is (0, 0), so this is the base point for the
linearization. Then, setting F (y, y 0) = (y 0)2 sin y + ey − 1, we have:
2
∂F
= (y 0)2 cos y + ey ,
∂y
∂F
= 2y 0 sin y.
∂y 0
Therefore,
∂F
(0, 0)
∂y
= 1, and
∂F
(0, 0)
∂y 0
= 0, and the linearized equation is
y 00 = y.
3d) The characteristic equation of the linearized equation with constant coefficients found in 3c is r 2 − 1 = 0. This equation has two real roots ±1.
Therefore, the general solution is
C1 et + C2 e−t
where C1 and C2 are arbitrary constants.
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