solutions to math 53 second exam | november 4,...

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.

Math 53, Autumn 2014 Solutions to Second Exam — November 4, 2014 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

)


(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.)



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)

).


(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.



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

)).


(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.


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.

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