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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).

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).

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.4: Multiple-angle Formulas
Problem 41E
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Check the errors in the computation below. Use the table for the answers.

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
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
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