solutions to math 53 second exam | november 4,...
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Solutions to Math 53 Second Exam — November 4, 2014
1. (20 points) (a) Find the maximal interval of definition of the solution to the initial value problem
ln
(t
2− 1
)y′ = y + e2t, y(3) = 1.
This is a linear equation, so we use Theorem 2.3.1 from the book. Put into standard form:
y′ −[
1
ln(t/2− 1)
]︸ ︷︷ ︸
p(t)
y =e2t
ln(t/2− 1)︸ ︷︷ ︸g(t)
.
The maximal interval of definition is the maximal open interval I containing t = 3 on whichboth p(t) and g(t) are continuous. For p(t), we require that t/2− 1 > 0, i.e., t > 2, so that thelogarithm is defined; furthermore, we need that t/2− 1 6= 1, i.e., t 6= 4, since otherwise we havedivision by zero. The same clearly holds for g(t). Therefore, p(t) and g(t) are simultaneouslycontinuous on 2 < t < 4 and t > 4. The initial condition at t = 3 hence gives I = (2, 4).
(b) Find all values (t0, y0) such that a unique solution is guaranteed for the initial value problem
y′ = |y − t|, y(t0) = y0.
(Hint: break into different cases.)
Solution 1: This is a nonlinear equation, so we use Theorem 2.3.2 from the book. Then
f(t, y) = |y − t|, ∂f
∂y=
{+1 if y > t,
−1 if y < t.
Clearly, f is continuous everywhere in the ty-plane, while fy is continuous in each open half-space y > t and y < t; however, it is discontinuous (in fact, undefined) at y = t. Therefore,the theorem applies only within each half-space and uniqueness (in some local neighborhood) isguaranteed for all (t0, y0) such that y0 6= t0.
Solution 2: Alternatively, we can write this as
y′ =
{y − t in R1 = {(t, y) | y ≥ t},t− y in R2 = {(t, y) | y ≤ t}
and apply the linear theorem to each case. Fix y0. Then if t0 < y0, applying the theorem inR1 gives uniqueness for all t < y; similarly, if t0 > y0, then applying the theorem in R2 givesuniqueness for all t > y. Note that we have excluded t0 = y0 since there is no open intervalcontaining it in either case. Therefore, uniqueness is guaranteed for all t0 6= y0.
Advanced: Actually, the problem is quite subtle and there is a unique solution for all (t0, y0). Solve each equationin R1,2 above and impose the initial condition y0 = t0 to get
y1 = t + 1− et−t0 in R1,
y2 = t− 1 + e−(t−t0) in R2,
where yi is the solution in Ri. Check that y1 is valid (i.e., satisfies y1 ≥ t) on t ≤ t0; similarly, y2 is valid ont ≥ t0. So y1,2 overlap only at t = t0, where, in fact, they both agree: y1,2(t0) = y0 and y′1,2(t0) = |y0 − t0| = 0.Thus, the solution is unique and can be “glued together” as
y =
{y1 if t ≤ t0,
y2 if t > t0.
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Math 53, Autumn 2014 Solutions to Second Exam — November 4, 2014 Page 2 of 8
2. (20 points) Consider the system of differential equations
x′ = Ax, A =
[2 α1 0
].
(a) Find the eigenvalues of A. For what values of α is the origin a saddle point?
The characteristic polynomial is det(A−λI) = λ2−Tr(A)λ+ det(A) = (λ−λ1)(λ−λ2). For ahomogeneous system to have a saddle point, it is necessary and sufficient to have real eigenvalueswith opposite signs. Hence, the system has a saddle point if and only if it has real roots andλ1λ2 = det(A) < 0. Thus the discriminant has to be positive, Tr(A)2 − 4 det(A) = 4 + 4α > 0,and the determinant has to be negative, det(A) = −α < 0. Hence, to have a saddle point, weneed to have α > 0.
(b) Let α = 3. Find the general solution.
For α = 3, the roots of the characteristic polynomial are λ1 = 3 and λ2 = −1. Suppose
v1 =
(s1s2
)is an eigenvector for λ1. Hence, we have
(−1 31 −3
)(s1s2
)=
(00
)
which implies −s1 + 3s2 = 0, so we can choose v1 =
(31
)to be an eigenvector for λ1. Now
assume v2 =
(r1r2
)is an eigenvector for λ1 = −1. To find v2, we need to solve the following
(3 31 1
)(r1r2
)=
(00
)
which implies r1 + r2 = 0, so we can choose v2 =
(1−1
)to be an eigenvector for λ2. The general
solution in the case of real different eigenvalues, is given by x(t) = c1eλ1tv1 + c2e
λ2tv2:
x(t) = c1e3t
(31
)+ c2e
−t(
1−1
)
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Math 53, Autumn 2014 Solutions to Second Exam — November 4, 2014 Page 3 of 8
(c) Let α = 3. For ease of reference, the corresponding system is
x′ =
[2 31 0
]x.
Draw a phase portrait that clearly shows the asymptotic behavior as t→∞. (Hint: it mayhelp to draw a few slope vectors.)
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Math 53, Autumn 2014 Solutions to Second Exam — November 4, 2014 Page 4 of 8
3. (20 points) Consider the system of differential equations
x′(t) = −yy′(t) = 4x.
(a) Find the general real-valued solution.
We begin by writing the equation in matrix form: Let ~z :=
(xy
), and we get
~z′ =
(0 −14 0
)~z.
Then we compute the characteristic polynomial in order to find the eigenvalues:
det(A− λI) =
∣∣∣∣−λ −14 −λ
∣∣∣∣ = λ2 + 4 = 0.
Thus the eigenvalues are ±2i. Picking one at random, say λ = 2i, we find an eigenvector, bysolving
(A− λI)v = 0,(−2i −1
4 −2i
)~v = 0.
We may choose our eigenvector to be
(1−2i
). The other eigenvector can then be chosen to be
the complex conjugate
(12i
).
The general solution, in complex form, is thus given by
~z(t) = c1e2it
(1−2i
)+ c2e
−2it(
12i
).
To find the real-valued solutions, we expand the first term using Euler’s formula:
e2it(
1−2i
)=
(cos(2t) + i sin(2t)−2i cos(2t) + 2 sin(2t)
)
=
(cos(2t)
2 sin(2t)
)+ i
(sin(2t)−2 cos(2t)
).
Thus, we can write down the general real solution:
~z(t) = c1
(cos(2t)
2 sin(2t)
)+ c2
(sin(2t)−2 cos(2t)
).
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(b) Find the particular solution passing through x(0) = 1, y(0) = 0. Draw a component plot.
Plugging in t = 0 to the general form, we have that(x(0)y(0)
)= c1
(10
)+ c2
(0−2
).
Equating components gives 1 = x(0) = c1 and 0 = y(0) = −2c2, so c1 = 1, c2 = 0, and thespecific solution is given by
x(t) = cos(2t), y(t) = 2 sin(2t).
(c) Identify the type of critical point at the origin. If it is a center or a spiral point, state the directionof rotation (clockwise or counterclockwise) in the xy-plane.
The node at the origin is a center , since the eigenvalues are complex and have 0 real part.
Evaluating the direction field at an arbitrary point, say
(10
), we find that
A
(10
)=
(04
),
so the center turns in the counterclockwise direction.
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4. (24 points) Consider the system of differential equations
x′ =
[0 1−4 4
]x.
(a) Find the general solution.
Let
A =
(0 1−4 4
).
Then
det(A− λI) = det
(−λ 1−4 4− λ
)= (λ− 2)2,
and A has only one eigenvalue λ = 2. Let
v =
(v1v2
).
Solving (0 1−4 4
)(v1v2
)= 2
(v1v2
),
we have
v =
(12
).
Since λ = 2,
A− λI =
(−2 1−4 2
).
Let
w =
(w1
w2
).
Solving (−2 1−4 2
)(w1
w2
)=
(12
),
we have
w =
(01
).
Therefore, we have
x1(t) = e2t(
12
),
and
x2(t) = te2t(
12
)+ e2t
(01
).
The general solution of the system is
x(t) = c1e2t
(12
)+ c2
(te2t
(12
)+ e2t
(01
)).
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(b) Draw a phase portrait that clearly shows the asymptotic behavior as t→∞. (Hint: it mayhelp to draw a few slope vectors.)
(c) Identify the type of critical point at the origin and its stability.
The origin (0, 0) is an unstable improper node.
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5. (16 points) (a) Rewrite the second-order linear differential equation
x′′ + 2x′ − etx = 1
as a first-order system of two equations.
Let’s introduce two new variables:x1 = x,
x2 = x′.
Then the equation x′′ + 2x′ − etx = 1 can be written as{x′1 = x2
x′2 + 2x2 − etx1 = 1, i.e.,
{x′1 = x2
x′2 = etx1 − 2x2 + 1.
So the system of equations is
d
dt
(x1x2
)=
(0 1et −2
)(x1x2
)+
(01
).
(b) Without computing eigenvalues/eigenvectors, show that
x1(t) = e2t[21
], x2(t) = e3t
[11
]form a fundamental pair of solutions for
x′(t) =
[1 2−1 4
]x.
First we need check ~x1 and ~x2 are solutions to the equation.
~x1′ = 2e2t
(21
)= e2t
(42
), A~x1 = e2t
(1 2−1 4
)(21
)= e2t
(42
),
~x2′ = 3e3t
(11
)= e3t
(33
), A~x1 = e3t
(1 2−1 4
)(11
)= e3t
(33
).
We find that~x1′ = A~x1, ~x2
′ = A~x2.
So ~x1 and ~x2 are solutions to the equation ~x ′ = A~x. (Notice that the problem requires notfinding eigenvalues and eigenvectors, we should check them directly.) To see that ~x1 and ~x2 forma fundamental set of solutions, we need compute the Wronskian:
W [~x1, ~x2](t) = det
(2e2t e3t
e2t e3t
)= e5t 6= 0.
Therefore, ~x1 and ~x2 form a fundamental set of solutions.