Download Problem Score Points for the Problem 1(a) 10 1(b) 10 2 20 3(a) 10 3

MIDTERM EXAM II (A)
110.302 DIFFERENTIAL EQUATIONS–SPRING 2014
APRIL 14, 2014
Instructions: The exam is 8 pages long, including this title page. The number
of points each problems is worth is listed after the problem number. The exam
totals to 100 points. For each item, please show your work or explain how you
reach your solutions. Please do all the work you wish graded on the exam paper.
Please indicate if you need to use the back of the exam paper. Good luck!
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Problem Score Points for the Problem
1(a)
10
1(b)
10
2
20
3(a)
10
3(b)
10
4
20
5
20
Total
100
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material or device.
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Date:
1
2
MIDTERM EXAM II (A), MATH302 DIFFERENTIAL EQUATIONS
Problem 1. (a) (10 pts) Consider the initial value problem
y 00 − 6y 0 + 9y = 0,
y(0) = 2, y 0 (0) = b.
Find the solution as a function of b, and then determine the critical value of b that
separates solutions that grow positively from those that eventually grow negatively.
Answers. The characteristic equation is
r2 − 6r + 9 = 0.
So it has a repeated real root r = 3 and the general solution of the differential
equation is given by
y = c1 e3t + c2 te3t .
Plugging the initial condition into the equation gives that
2 = c1
and
3c1 + c2 = b.
Thus
c1 = 2 and c2 = b − 6.
Hence, the solution of the initial value problem is given by
y = 2e3t + (b − 6)te3t .
Finally, if b ≥ 6, the solution, as t → ∞, is dominated by (b − 6)te3t , growing
positively if b > 6 while negatively if b < 6. Therefore, b = 6 is the critical value of
b that separates the asymptotic behavior of the solutions.
MIDTERM EXAM II (A), MATH302 DIFFERENTIAL EQUATIONS
3
(b) (10 pts) Solve the differential equation
y 00 − 6y 0 + 9y = te3t .
Answers. Since 3 is a root of the characteristic equation with multiplicity two, the
particular solution Y (t) should be of the following form
Y (t) = t2 (At + B)e3t .
Then lets calculate the derivatives of Y (t) as follows.
Y 0 (t) = (3At2 + 2Bt + 3At3 + 3Bt2 )e3t ,
and
Y 00 (t) = (6At + 2B + 18At2 + 12Bt + 9At3 + 9Bt2 )e3t .
Plugging these into equation gives that
6At + 2B = t,
which implies that A = 1/6 and B = 0.
Therefore, the general solution is given by
1
y = t3 e3t + c1 e3t + c2 te3t .
6
4
MIDTERM EXAM II (A), MATH302 DIFFERENTIAL EQUATIONS
Problem 2. (20 pts) Find the general solution of the differential equation
y 00 − 4y 0 + 4y =
e2t
.
t2 + 1
Answers. The characteristic equation is
r2 − 4r + 4 = 0,
having a repeated real root r = 2. Thus
y1 (t) = e2t
and y2 (t) = te2t
are a fundamental set of solutions of the corresponding homogeneous equation.
Furthermore, the Wronskian of y1 and y2 is
W [y1 , y2 ] = e4t .
Using the variation of parameters, the general solution is given by
Z
Z
1
t
2t
dt
+
te
dt
y(t) = −e2t
t2 + 1
t2 + 1
1
= − e2t log(t2 + 1) + te2t arctan t + c1 e2t + c2 te2t .
2
MIDTERM EXAM II (A), MATH302 DIFFERENTIAL EQUATIONS
5
Problem 3. (a) (10 pts) Solve the differential equation
y 000 − y 00 + y 0 − y = 0.
Answers. The characteristic equation is
r3 − r2 + r − 1 = 0.
The roots are r1 = 1, r2 = i, and r3 = −i.
Hence the general solution is given by
y(t) = c1 et + c2 cos t + c3 sin t.
6
MIDTERM EXAM II (A), MATH302 DIFFERENTIAL EQUATIONS
(b) (10 pts) Solve the differential equation
y 000 − y 00 + y 0 − y = cos(2t).
Answers. The particular solution Y (t) is of the following form
Y (t) = A cos(2t) + B sin(2t).
Then we calculate the derivatives of Y (t) as follows.
Y 0 (t) = −2A sin(2t) + 2B cos(2t),
Y 00 (t) = −4A cos(2t) − 4B sin(2t),
and
Y 000 (t) = 8A sin(2t) − 8B cos(2t).
Plugging into the equation implies that
(8A + 4B − 2A − B) sin(2t) + (−8B + 4A + 2B − A) cos(2t) = cos(2t).
Thus,
6A + 3B = 0 and 3A − 6B = 1.
Hence A = 1/15 and B = −2/15. The general solution is given by
1
2
y(t) =
cos(2t) −
sin(2t) + c1 et + c2 cos t + c3 sin t.
15
15
MIDTERM EXAM II (A), MATH302 DIFFERENTIAL EQUATIONS
7
Problem 4. (20 pts) Given solutions y1 (t) = t2 and y2 (t) = t3 , solve the
following equation
t2 (t + 3)y 000 − 3t(t + 2)y 00 + 6(1 + t)y 0 − 6y = 0,
t > 0.
Answers. It is easy to observe that y3 (t) = t + 1. Then calculate the Wronskian of
y1 , y2 , and y3 :
2
t
t3 t + 1 1 = 2t3 + 6t2 > 0
W [y1 , y2 , y3 ](t) = 2t 3t2
2 6t
0 for t > 0.
Therefore, the general solution is given by
y(t) = c1 t2 + c2 t3 + c3 (t + 1).
8
MIDTERM EXAM II (A), MATH302 DIFFERENTIAL EQUATIONS
Problem 5. (20 pts) Solve the initial value problem for the system
x01 = x1 − 2x2 ,
x02
= 3x1 − 4x2 ,
x1 (0) = −1
x2 (0) = 2
Answers. From the first equation,
1
1
x2 = (x1 − x01 ) and x02 = (x01 − x001 ).
2
2
Substituting them into the second equation,
1 0
(x − x001 ) = 3x1 − 2(x1 − x01 ).
2 1
Thus
x001 + 3x01 + 2x1 = 0,
and
x1 = c1 e−t + c2 e−2t .
Moreover,
3c2 −2t
e .
x2 = c1 e−t +
2
By the initial conditions,
3
−1 = c1 + c2 and 2 = c1 + c2 .
2
Thus c1 = −7 and c2 = 6.