MATH 215/255
Assignment 5
5.
Determine the general solution for each of the following differential equations,
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(a) y00 − 4y = xe2x
(b) y00 + 2y0 = 1 + e−2x
§8.1 §8.2 §8.3
(c) y00 + 4y = xsin(2x)
1. (§1.3) Solve y0 + y2 sinx = 0.
Answer. If y(x0) = 0 for some x0, then y(x) = 0 for all x. If y 6= 0, the DE may be written as y−2dy = −sinxdx.
Integrating both sides, we get −y−1 = cosx + c, or y = −1/(cosx + c).
6.
Note: For any given initial data y(x0) = y0 6= 0, we have cosx + c 6= 0 for x near x0. The interval of
existence is Rinite if |c| ≤ 1.
2. (§1.4) Solve y0 − y = 2te2t,
Answer.
Let
. Find f(t) = L−1[F(s)].
Answer.
.
Using the Rirst shifting property and
, we obtain:
y(0) = 1.
.
µ(t) = e−t so that (e−ty)0 = 2tet and thus
Z
e−ty(t) =
Z
2tetdt = 2tet − 2
7.
etdt = 2(t − 1)et + c.
Let
. Find f(t) = L−1[F(s)].
Answer.
Setting t = 0 and y = 1 we get 1 = −2 + c, or c = 3. Hence y(t) = 2(t − 1)e2t + 3et.
3.
Notice that:
We Rirst rewrite:
.
Using the Rirst shifting property and
f(t) = L−1[F(s)](t) = 1 − e−t − te−t = 1 − e−t [1 + t]
Find the general solution of the following differential equations. Determine the complementary and
particular parts of the solutions.
(a) ty0 − y − 2te2t = 0,
, we obtain:
8.
Find the inverse Laplace transform of each of the following functions,
(b) y0 + (2/t)y − cos(t)/t2 = 0,
(c) y0 − 10y + cos(t)e2t = 0
4.
Find the solution of the initial value problem,
Answer.
The general equation is given by,
,
.
Using the initial conditions, we see that:
,
9.
must be the particular solution to the IVP.
In each of the following problems, solve the initial value problem, and explain the long-term
behaviour of solutions (i.e. trajectories) as t → ∞.
1
2
(1)
(b) Show that for any trajectory except the origin, the distance from the origin is converging tozero. Conclude that
the origin is asymptotically stable.
Hints: i) Consider f(t) = 3(x(t))2 + (y(t))2 and show it has negative derivative.
(2)
ii) It is indeed a spiral sink.
Answer.
Answer.
The solutions to each of the two systems are given by:
and
(a) For critical points,
y = (x2 + y2)x,
One solution is the origin. If (x,y) 6= (0,0), then x2 + y2 > 0, and from
3x = −(x2 + y2)y.
(3)
we get y = 0. Back to (??), we get x = 0. Thus (0,0) is the only Eixed point.
The Jacobian matrix at (0,0) is the matrix
. The linearization is
x0 = y,
10. For the following systems, Rind the linearization at (0,0).
y0 = −3x.
√
The eigenvalues of J satisfy λ2 +3 = 0, thus λ1,2 = ± 3i. Since Re λ1,2 = 0, the Eixed point is a (linear) center. We cannot tell the behaviour of
solutions near a critical point whose linearization is a centre.
(b) We have
f0(t) = 6xx0 + 2yy0 = 6x(y − (x2 + y2)x) + 2y(−3x − (x2 + y2)y).
Noting the cancellation of 6xy, we get
,
Answer.
system
f0(t) = −(6x2 + 2y2)(x2 + y2),
In each case it is enough to compute the Jacobian at (0,0), since the linearization is the
f0(t) < 0
unless (x,y) = (0,0).
Hence f(t) is decreasing to 0. As f(t) ≥ (x(t))2 + (y(t))2, the square of the distance to the origin, the distance to the origin is converging to
zero, and all solutions are converging to (0,0). Hence the Eixed point (0,0) is asymptotically stable.
13.
Consider the competition system (x for coyotes and y for foxes)
x0 = 3x − xy − 2x2,
y0 = 3y − xy − 2y2.
Find all Eixed points. Denote by R the unique Eixed point (x0,y0) with x0 > 0 and y0 > 0. Is it stable (co-existence) or unstable (one
species extinction)? Answer.
.
11. (8.2.7) Consider the system: x0 = f(x,y), y0 = g(x,y). Suppose that g(x,y) > 1 for all (x,y). Are there any
critical points? What can we say about the trajectories as t goes to inRinity?
Answer.
There is no critical point since g(x,y) never vanishes.
The derivative of the y-coordinate of solutions (x(t),y(t)) is always larger than 1, therefore y(t) → +∞ as
t → ∞.
Critical points satisfy
f = x(3 − y − 2x) = 0,
g = y(3 − x − 2y) = 0.
If x = 0, then y = 0 or y = 3/2.
If x 6= 0, then 3 − y − 2x = 0. Either y = 0 (implying x = 3/2) or 3 − y − 2x = 0 = 3 − x − 2y, resulting in x = y = 1.
Hence we have 4 Dixed points: (0,0), (0,3/2), (3/2,0), and (1,1).
We have R = (1,1).
The Jacobian matrix is
.
At (1,1),
12. Consider the system
x0 = y − (x2 + y2)x,
.
y0 = −3x − (x2 + y2)y.
(a) Find ALL critical points and their linearization. Can you tell the behaviour of the solutionsnear
critical points from the linearization?
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The linearization is a sink. The nonlinear system near R is also a sink. It is asymptotically stable.
Remarks. (i) This model is co-existence: most solutions converge to (1,1).
(ii) The coefEicients of xy measure inter-species competition. In this problem they are −1, weaker than −2 of the previous problem.
Hence there is more chance for co-existence.
4