ERROR CORRECTION TEST ITEM: DIRECTIONS: The entire test has been answered for you. Your job is to check the answers for errors. If there are errors, do the following in the table after the test sample: 1. Correct the answer. 2. Explain why the answer is wrong. 3. Then explain why your correction is right. ANSWERED TEST WITH ERRORS: Use your knowledge on Higher-order derivative and Chain Rule to show that z" + 4z' + 8z = 0 if z = e-2x (sin 2x + cos 2x). -2x (sin 2x + cos 2x) Z = e Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z' = e-"Dx(sin u + cos u) + (sin u + cos u)Dx(e¯u) z'%3Dе-"(сos и — sin u) — е-"(sinu + cosuи) = e-u cos u - e-u sin u - e-u sin u - e-u cos u dy = -2e-u sinu du dy du Substitute = -2e-" sin u, = 2 using the Chain Rule, dx du dz dy du (-2e-u sin u)2 = -4e-" sin u %3D %3D du du dx Substitute u = 2x, z' = -4e-2x sin 2x To get the second derivative, again let u = 2x then du = 2dx du = 2 dx z' = -4e-u sin u z" = -4[e-uDx(sin u) + sin u Dz(e-") = -4e-" cos u – 4e-u sin u dy = -4e -u cos u – 4e-u sin u du dy Substitute du du 4e-" cos u – 4e-u sin u, = 2 using the Chain Rule, dx dz dy du 3D (-4е и соs и — 4e-u sin u)2 — —8e-и cos и - 8e- sin u du du dx WBL S-OBE MELC-Aligned Self Learning Module Basic Calculus 22 Substitute u = 2x, z" = -8e-2x cos 2x – 8e-2x sin 2x Show z" + 4z' + 8z = 0. (-4e-2x sin 2x) + 4(-8e-2x cos 2x – 8e-2x sin 2x) + 8[e-2x (sin 2x + cos 2x)] = 0 -4e-2x sin 2x – 32e-2x cos 2x – 32e-2x sin 2x + 8e-2x sin 2x + 8e-2x cos 2x = 0 -24e-2x cos 2x – 28e-2x sin 2x + 0 IDENTIFIED ERROR CORRECTION OF ERROR EXPLANATION OF CORRECTION
ERROR CORRECTION TEST ITEM: DIRECTIONS: The entire test has been answered for you. Your job is to check the answers for errors. If there are errors, do the following in the table after the test sample: 1. Correct the answer. 2. Explain why the answer is wrong. 3. Then explain why your correction is right. ANSWERED TEST WITH ERRORS: Use your knowledge on Higher-order derivative and Chain Rule to show that z" + 4z' + 8z = 0 if z = e-2x (sin 2x + cos 2x). -2x (sin 2x + cos 2x) Z = e Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z' = e-"Dx(sin u + cos u) + (sin u + cos u)Dx(e¯u) z'%3Dе-"(сos и — sin u) — е-"(sinu + cosuи) = e-u cos u - e-u sin u - e-u sin u - e-u cos u dy = -2e-u sinu du dy du Substitute = -2e-" sin u, = 2 using the Chain Rule, dx du dz dy du (-2e-u sin u)2 = -4e-" sin u %3D %3D du du dx Substitute u = 2x, z' = -4e-2x sin 2x To get the second derivative, again let u = 2x then du = 2dx du = 2 dx z' = -4e-u sin u z" = -4[e-uDx(sin u) + sin u Dz(e-") = -4e-" cos u – 4e-u sin u dy = -4e -u cos u – 4e-u sin u du dy Substitute du du 4e-" cos u – 4e-u sin u, = 2 using the Chain Rule, dx dz dy du 3D (-4е и соs и — 4e-u sin u)2 — —8e-и cos и - 8e- sin u du du dx WBL S-OBE MELC-Aligned Self Learning Module Basic Calculus 22 Substitute u = 2x, z" = -8e-2x cos 2x – 8e-2x sin 2x Show z" + 4z' + 8z = 0. (-4e-2x sin 2x) + 4(-8e-2x cos 2x – 8e-2x sin 2x) + 8[e-2x (sin 2x + cos 2x)] = 0 -4e-2x sin 2x – 32e-2x cos 2x – 32e-2x sin 2x + 8e-2x sin 2x + 8e-2x cos 2x = 0 -24e-2x cos 2x – 28e-2x sin 2x + 0 IDENTIFIED ERROR CORRECTION OF ERROR EXPLANATION OF CORRECTION
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Identified the error and correct then explain
![ERROR CORRECTION TEST ITEM:
DIRECTIONS: The entire test has been answered for you. Your job is to check the answers
for errors. If there are errors, do the following in the table after the test sample:
1. Correct the answer.
2. Explain why the answer is wrong.
3. Then explain why your correction is right.
ANSWERED TEST WITH ERRORS:
Use your knowledge on Higher-order derivative and Chain Rule to show that
z" + 4z' + 8z = 0 if z = e-2x (sin 2x + cos 2x).
-2x (sin 2x + cos 2x)
Z = e
Let u = 2x, then
du = 2dx
du
= 2
dx
Z = e-"(sin u + cos u)
z'%3D е"Dx(siп и + сos u) + (sin u + cos u)Dx(e"")
z'%3Dе-"(сos и - sin u) — е-" (sinu + cosuи)
в е-и сos u — е-и sin u — е-И sin u — е-И сos u
dy
— -2е и sin u
du
dy
— -2е И sinu,
du
du
Substitute
= 2 using the Chain Rule,
dx
dz
dy du
(-2е-u sin u)2 3D —4е-и sin и
%3D
%3D
du
du dx
Substitute u = 2x,
z' = -4e-2x sin 2x
To get the second derivative, again let u = 2x then
du = 2dx
du
= 2
dx
z' = -4e-u sin u
z" = -4[e-uDx(sin u) + sin u Dz(e-")
3 -4e " сos и — 4е И sin и
dy
— —4е и соs u — 4e-u sin u
du
dy
Substitute
du
du
4e-" cos u – 4e-u sin u,
= 2 using the Chain Rule,
dx
dz
dy du
3D (-4e и соs и — 4e-u sin u)2 — —8e-и cos u — 8e-u sin u
du
du dx
WBLS-OBE
MELC-Aligned
Self-Learning Module
Basic Calculus
22
Substitute u = 2x,
z" = -8e-2x cos 2x – 8e-2x sin 2x
Show z" + 4z' + 8z = 0.
(-4e-2x sin 2x) + 4(-8e-2x cos 2x – 8e-2x sin 2x) + 8[e-2x (sin 2x + cos 2x)] = 0
-4e-2x sin 2x – 32e-2x cos 2x – 32e-2x sin 2x + 8e-2x sin 2x + 8e-2x cos 2x = 0
-24e-2x cos 2x – 28e-2x sin 2x + 0
IDENTIFIED ERROR
CORRECTION OF ERROR
EXPLANATION OF
CORRECTION](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F061d3272-0af8-45de-8848-27f401374c63%2F5853109b-86a0-4c86-bd97-fd0e3f214065%2F22duog_processed.jpeg&w=3840&q=75)
Transcribed Image Text:ERROR CORRECTION TEST ITEM:
DIRECTIONS: The entire test has been answered for you. Your job is to check the answers
for errors. If there are errors, do the following in the table after the test sample:
1. Correct the answer.
2. Explain why the answer is wrong.
3. Then explain why your correction is right.
ANSWERED TEST WITH ERRORS:
Use your knowledge on Higher-order derivative and Chain Rule to show that
z" + 4z' + 8z = 0 if z = e-2x (sin 2x + cos 2x).
-2x (sin 2x + cos 2x)
Z = e
Let u = 2x, then
du = 2dx
du
= 2
dx
Z = e-"(sin u + cos u)
z'%3D е"Dx(siп и + сos u) + (sin u + cos u)Dx(e"")
z'%3Dе-"(сos и - sin u) — е-" (sinu + cosuи)
в е-и сos u — е-и sin u — е-И sin u — е-И сos u
dy
— -2е и sin u
du
dy
— -2е И sinu,
du
du
Substitute
= 2 using the Chain Rule,
dx
dz
dy du
(-2е-u sin u)2 3D —4е-и sin и
%3D
%3D
du
du dx
Substitute u = 2x,
z' = -4e-2x sin 2x
To get the second derivative, again let u = 2x then
du = 2dx
du
= 2
dx
z' = -4e-u sin u
z" = -4[e-uDx(sin u) + sin u Dz(e-")
3 -4e " сos и — 4е И sin и
dy
— —4е и соs u — 4e-u sin u
du
dy
Substitute
du
du
4e-" cos u – 4e-u sin u,
= 2 using the Chain Rule,
dx
dz
dy du
3D (-4e и соs и — 4e-u sin u)2 — —8e-и cos u — 8e-u sin u
du
du dx
WBLS-OBE
MELC-Aligned
Self-Learning Module
Basic Calculus
22
Substitute u = 2x,
z" = -8e-2x cos 2x – 8e-2x sin 2x
Show z" + 4z' + 8z = 0.
(-4e-2x sin 2x) + 4(-8e-2x cos 2x – 8e-2x sin 2x) + 8[e-2x (sin 2x + cos 2x)] = 0
-4e-2x sin 2x – 32e-2x cos 2x – 32e-2x sin 2x + 8e-2x sin 2x + 8e-2x cos 2x = 0
-24e-2x cos 2x – 28e-2x sin 2x + 0
IDENTIFIED ERROR
CORRECTION OF ERROR
EXPLANATION OF
CORRECTION
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