Homework Help is Here – Start Your Trial Now!
Math
Calculus
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). Z = e-2x (sin 2x + cos 2x) Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z'%3Dе"Dx(sin и + cos u) + (sin и + cos u)Dx(е ") z'%3Dе " (сos u — sin u) — е (sinu + cosu) coS и — е "и sin u — e"u sin u — e"и cos u dy = -2e¬u sin u du dy = -2e¬u sin u, du du Substitute = 2 using the Chain Rule, dx dz dy du (-2e¬" sin u)2 = -4e¬u sin u 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 -4[e-"Dx(sin u) + sin u Dx(e¬u) 3 -4е И cos и — 4е -И sin u z" = dy — — 4е и соs и — 4e-U sin u du dy = -4e-u du Substitute cOS и — 4e"u sin u, = 2 using the Chain Rule, dx du dz dy du (-4е " cos u — 4е И sin u)2 3D —8е И сosu — 8e u sin u du du dx

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). Z = e-2x (sin 2x + cos 2x) Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z'%3Dе"Dx(sin и + cos u) + (sin и + cos u)Dx(е ") z'%3Dе " (сos u — sin u) — е (sinu + cosu) coS и — е "и sin u — e"u sin u — e"и cos u dy = -2e¬u sin u du dy = -2e¬u sin u, du du Substitute = 2 using the Chain Rule, dx dz dy du (-2e¬" sin u)2 = -4e¬u sin u 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 -4[e-"Dx(sin u) + sin u Dx(e¬u) 3 -4е И cos и — 4е -И sin u z" = dy — — 4е и соs и — 4e-U sin u du dy = -4e-u du Substitute cOS и — 4e"u sin u, = 2 using the Chain Rule, dx du dz dy du (-4е " cos u — 4е И sin u)2 3D —8е И сosu — 8e u sin u du du dx

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
See similar textbooks
icon
Related questions
Question

Please check the errors in the computation below. Insert the answer in the following table. 

Substitute u = 2x, z" 3D —8е2х cos 2x — 8е 2x sin 2х 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 — 28е- e-2x sin 2x 0 IDENTIFIED ERROR CORRECTION OF ERROR EXPLANATION OF CORRECTION
Transcribed Image Text:Substitute u = 2x, z" 3D —8е2х cos 2x — 8е 2x sin 2х 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 — 28е- e-2x sin 2x 0 IDENTIFIED ERROR CORRECTION OF ERROR EXPLANATION OF CORRECTION
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). Z = e-2x (sin 2x + cos 2x) Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z'%3Dе"Dx(sin и + cos u) + (sin и + cos u)Dx(е ") z'%3Dе " (сos u — sin u) — е (sinu + cosu) coS и — е "и sin u — e"u sin u — e"и cos u dy = -2e¬u sin u du dy = -2e¬u sin u, du du Substitute = 2 using the Chain Rule, dx dz dy du (-2e¬" sin u)2 = -4e¬u sin u 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 -4[e-"Dx(sin u) + sin u Dx(e¬u) 3 -4е И cos и — 4е -И sin u z" = dy — — 4е и соs и — 4e-U sin u du dy = -4e-u du Substitute cOS и — 4e"u sin u, = 2 using the Chain Rule, dx du dz dy du (-4е " cos u — 4е И sin u)2 3D —8е И сosu — 8e u sin u du du dx
Transcribed Image Text: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). Z = e-2x (sin 2x + cos 2x) Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z'%3Dе"Dx(sin и + cos u) + (sin и + cos u)Dx(е ") z'%3Dе " (сos u — sin u) — е (sinu + cosu) coS и — е "и sin u — e"u sin u — e"и cos u dy = -2e¬u sin u du dy = -2e¬u sin u, du du Substitute = 2 using the Chain Rule, dx dz dy du (-2e¬" sin u)2 = -4e¬u sin u 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 -4[e-"Dx(sin u) + sin u Dx(e¬u) 3 -4е И cos и — 4е -И sin u z" = dy — — 4е и соs и — 4e-U sin u du dy = -4e-u du Substitute cOS и — 4e"u sin u, = 2 using the Chain Rule, dx du dz dy du (-4е " cos u — 4е И sin u)2 3D —8е И сosu — 8e u sin u du du dx
Expert Solution
steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS