Download Instructors Solution Key All Sections Problem Mark Total 1 16 2 14 3

Math 202- Elements of Differential Equations
Major Exam II- Semester 102 (2010-2011)
King Fahd University of Petroleum and Minerals
Department of Mathematics and Statistics
Duration: 2h
Instructors Solution Key
All Sections
Problem Mark Total
1
16
2
14
3
14
4
14
5
14
6
16
Total
88
1
Exercise 1 (16 points)
a.) Find A, B and C such that y = Ax2 + Bx + C is a solution to the differential
equation
y 00 + 4y 0 − y = −x2 + 1.
b.) Find D and E such that y = (Dx + E)e−x is a solution to the differential equation
y 00 + 4y 0 − y = xe−x − e−x .
c.) Find a particular to the following differential equation
y 00 + 4y 0 − y = −x2 + 1 + xe−x − e−x .
Solution
a)y = Ax2 + Bx + C, y 0 = 2Ax + B, y 00 = 2A. ⇒ 1 point
Then
y 00 + 4y 0 − y = 2A + 4(2Ax + B) − (Ax2 + Bx + C)
= −Ax2 + (8A − B)x + 2A + 4B − C .
= −x2 + 1.
⇒ 2 points

= −1
 −A
8A − B
=
0 Thus A = 1, B = 8 and C = 33.
By identification we have

2A + 4B − C =
1
⇒ 3 points
1.5 points if the system is correct but A, B and C are wrong.
The maximum number of points here is 6.
b) y = (Dx + E)e−x , y 0 = (−Dx − E + D)e−x , y 00 = (Dx + E − 2D)e−x . ⇒ 2 points
Then
y 00 + 4y 0 − y = (Dx + E − 2D + 4(−Dx − E + D) − (Dx + E))e−x
⇒ 2 points
= (−4Dx − 4E + 2D)e−x .
By identification we have −4D = 1 and 2D − 4E = −1. Thus D = − 41 and E = 81 .
⇒ 2 points
The maximum number of points here is 6.
c) From a) and b) we deduce that yp = x2 + 8x + 33 − 14 xe−x + 81 e−x .
⇒ 2 points for each part.
Give 4 points if the student solves the equation without using a) and b) and
gets the correct answer.
The maximum number of points here is 4.
2
Exercise 2 (14 points)
a) Verify that y1 = x is a solution of
(1 + x2 )y 00 − 2xy 0 + 2y = 0.
b) Find the general solution of the differential equation.
Solution
a) y1 = x, y10 = 1, y100 = 0.
(1 + x2 )y100 − 2xy10 + 2y1 = −2x + 2x = 0.
Give 0 or 4 for this question; No partial credit if the students did not get
the correct answer.
The maximum number of points here is 4.
b) We first write the differential equation in its standard form
y 00 −
2x
y0
1+x2
+
2
y
1+x2
= 0. ⇒ 2 points
Let y2 be the second solution, then
R − R P (x)dx
y2 = y1 e y2 dx ⇒ 2 points
1
R
y2 = x
R
e
2x dx
1+x2
x2
dx = x
R
eln(1+x
x2
2)
dx = x
R
1+x2
dx
x2
⇒ 2 points (1+1 for the two last steps, 0 if the student did not integrate.)
R
Thus y2 = x (1 + x12 )dx = x(x − x1 ) = x2 − 1.
⇒ 2 points (Penalize (-1) just in one place if the student writes y2 = x2 + 1.)
The general solution is y = c1 y1 + c2 y2 = c1 x + c2 (x2 − 1), ⇒ 2 points
with c arbitrary constants.
If the student did not use the standard form: Give 2 points if he wrote
the correct formula for y2 and add 1 point if he wrote the form of the general
solution correctly.
The maximum number of points here is 10.
3
Exercise 3 (14 points)
What is the general solution of
(D2 − 4)(D − 2)2 (8D2 + 4D + 1)y = 0
Solution
First Approach:
(D2 − 4) = (D + 2)(D − 2) Thus e2x and e−2x are solutions. ⇒ 3 points
Since e2x was already found, (D − 2)2 suggests that xe2x and x2 e2x are solutions.
⇒ 4 points
For 8D2 + 4D + 1 the roots of the auxiliary equation are − 14 ± 41 i. ⇒ 3 points
x
x
Hence e− 4 cos( x4 ) and e− 4 sin( x4 ) are solutions. ⇒ 1 point
The general solution is
x
x
y = c1 e−2x + c2 e2x + c3 xe2x + c4 x2 e2x + c5 e− 4 cos( x4 ) + c6 e− 4 sin( x4 ), ⇒ 3 points
with c arbitrary constants.
The maximum number of points here is 14.
Second Approach:
The auxiliary equation is: (m2 − 4)(m − 2)2 (8m2 + 4m + 1) = 0 ⇒ 2 points
The roots are −2, 2 repeated 3 times and − 41 ± 14 i ⇒ 3 points, subtract 1 point if
the student did not give the conjugate complex root.
x
x
The solutions are e−2x , e2x , xe2x , x2 e2x , e− 4 cos( x4 ) and e− 4 sin( x4 ). ⇒ 6 points, 1
point for each solution.
The general solution is
x
x
y = c1 e−2x + c2 e2x + c3 xe2x + c4 x2 e2x + c5 e− 4 cos( x4 ) + c6 e− 4 sin( x4 ), ⇒ 3 points
with c arbitrary constants.
The maximum number of points here is 14.
Third Approach:
If the student gives directly the general solution without details: give full
mark if the answer is correct OR follow the point distribution for each solution
if the answer is not correct.
4
Exercise 4 (14 points)
Determine the form of a general solution for
y 000 − 2y 00 = 5x − x3 e2x + x2 e−x sin(2x)
(Do not find the constants values.)
Solution
The differential equation could be written: D2 (D − 2)y = 5x − x3 e2x + x2 e−x sin(2x).
Thus yc (x) = a + bx + ce2x , a, b and c are arbitrary constants. ⇒ 3 points
The annihilators of the right hand side are:
5x ⇒ D2 . ⇒ 1 point
x3 e2x ⇒ (D − 2)4 .⇒ 2 points
x2 e−x sin(2x) ⇒ (D2 + 2D + 5)3 . ⇒ 3 points
Thus (D2 + 2D + 5)3 D3 (D − 2)5 y = 0.
A particular solution would have the form:
yp (x) = c1 x2 + c2 x3 +
c3 xe2x + c4 x2 e2x + c5 x3 e2x + c6 x4 e2x +
⇒ 4 points
c7 e−x cos(2x) + c8 e−x sin(2x) + c9 xe−x cos(2x) + c10 xe−x sin(2x)+
c11 x2 e−x cos(2x) + c12 x2 e−x sin(2x).
Thus the general solution could be written
y(x) = a + bx + ce2x +
c1 x 2 + c2 x 3 +
c3 xe2x + c4 x2 e2x + c5 x3 e2x + c6 x4 e2x +
c7 e−x cos(2x) + c8 e−x sin(2x) + c9 xe−x cos(2x) + c10 xe−x sin(2x)+
c11 x2 e−x cos(2x) + c12 x2 e−x sin(2x).
⇒ 1 point (also give this 1 point if the students just writes y = yc + yp ).
The maximum number of points here is 14.
5
Exercise 5 (14 points)
Find a particular solution of the differential equation:
4y 00 + 16y = csc 2x.
Solution
We first write the standard form
y 00 + 4y = 14 csc 2x. ⇒ 1 point
The auxiliary equation of the homogeneous equation is m2 + 4 = 0. ⇒ 1 point
Thus yc (x) = c1 cos(2x) + c2 sin(2x), ⇒ 2 point
with c arbitrary constants.
We use the variation of parameters method to find yp .
cos(2x)
sin(2x)
= 2 6= 0. ⇒ 2 points
w = −2sin(2x) 2cos(2x) Let yp = u1 y1 + u2 y2 , then
0
sin(2x)
1
0
u1 = 2 1
csc(2x) 2cos(2x)
4
1
= (−sin(2x)csx(2x)) = − 1 .
8
8
⇒ 2 points, (1 point for w1 and 1 point for u01 ).
And
cos(2x)
u02 = 21 −2sin(2x)
1
0
= (cos(2x)csc(2x)) =
1
8
csc(2x)
4
1 cos(2x)
.
8 sin(2x)
⇒ 2 points, (1 point for w2 and 1 point for u02 ).
Integrating u01 and u02 yields:
u1 = − 81 x ⇒ 1 point
and u2 =
1
16
ln |sin(2x)|. ⇒ 2 points
Hence
yp (x) = − 18 xcos(2x) +
1
sin(2x) ln |sin(2x)|.
16
⇒ 1 point
If the student did not write the standard form: Penalize 3 points;
⇒ 1 point for the standard form, 1 point for w1 and 1 point w2 .
The maximum number of points here is 14.
6
Exercise 6 (16 points)
a) Solve the Cauchy-Euler differential equation: x2 y 00 − xy 0 + 2y = 0, x ∈ (0, ∞).
b) Convert the following equation to a Non-Homogeneous linear differential equation
with constant coefficients
x2 y 00 − 2xy 0 + 2y = ln(x), x ∈ (0, ∞).
(Do not solve the differential equation.)
Solution
a) Let y(x) = xm , y 0 (x) = mxm−1 and y 00 = m(m − 1)xm−2 .
The auxiliary equation is m2 − 2m + 2 = 0. ⇒ 4 points
Thus ∆ = −4 and m = 1 ± i. ⇒ 2 points
Hence y(x) = c1 xcos(ln x) + c2 xsin(ln x), ⇒ 2 points
with c arbitrary constants.
The maximum number of points here is 8.
b) To convert the equation to a Non-Homogeneous linear differential equation with
constant coefficients we assume that x = et or t = ln x. ⇒ 1 point
It follows that
dy
dx
=
And
dy dt
dt dx
d2 y
dx2
=
=
1 dy
.
x dt
1 d
x dx
⇒ 2 points
dy
dy
+ dt − x12 =
dt
1
x2
d2 y
dt2
−
dy
dt
. ⇒ 3 points
Substitution yields
2
− 2x x1 dy
+ 2y = t.
x2 x12 ddt2y − dy
dt
dt
⇒
d2 y
dt2
− 3 dy
+ 2y = t, t ∈ (−∞, ∞). ⇒ 2 points
dt
Subtract 0.5 point if the student did not write the interval of t.
The maximum number of points here is 8.
7