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Problems cach of Problems 1 through 8: Find the solution of the given initial value problem. Oh. Plot a graph of the solution. L+2y + 2y = 6(1 – ); y(0) = 1, y(0) =0 2 Y+4y= 6(t-#) - 6(1 – 2*); y(0) = 0, y(0) = 0 3Y+3y +2y = 6(1 - 5) + u 10(1); y(0) = 0, y(0) =1/2 4. y"+ 2y +3y = sin r +8(1- 3n); y(0) =0, y(0) =0 5. y+y=6(1-2r) cos r; y(0) -0, y(0) = 1 6. y"+4y = 26(1-/4); y(0) =0, y(0) =0 7. y+2y +2y cost + 6(t- /2): y(0) = 0, y(0) = 8. y4) - y = 6(t – 1); y(0) = 0, y(0) = 0, y"(0) = 0, y"(0) =0 %3D

Problems cach of Problems 1 through 8: Find the solution of the given initial value problem. Oh. Plot a graph of the solution. L+2y + 2y = 6(1 – ); y(0) = 1, y(0) =0 2 Y+4y= 6(t-#) - 6(1 – 2*); y(0) = 0, y(0) = 0 3Y+3y +2y = 6(1 - 5) + u 10(1); y(0) = 0, y(0) =1/2 4. y"+ 2y +3y = sin r +8(1- 3n); y(0) =0, y(0) =0 5. y+y=6(1-2r) cos r; y(0) -0, y(0) = 1 6. y"+4y = 26(1-/4); y(0) =0, y(0) =0 7. y+2y +2y cost + 6(t- /2): y(0) = 0, y(0) = 8. y4) - y = 6(t – 1); y(0) = 0, y(0) = 0, y"(0) = 0, y"(0) =0 %3D

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...
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1,2,4 Only A please

Problems cach of Problems 1 through 8: A Find the solution of the given initial value problem. O b. Plot a graph of the solution. 1+2y' +2y = 6(1-); y(0) = 1, y'(0) = 0 2 y +4y = 6(t – ) - 6(1 – 27); y(0) = 0, y'(0) = 0. 3 Y+3y +2y = 6(1 - 5) + uj0(1): y(0) = 0, y'(0) = 1/2 4. y"+ 2y + 3y = sinr + 6(1 - 3m); y(0) = 0, y'(0) =0 5. y"+ y = 6(1- 2x) cos t; y(0) - 0, y(0) = 1 6. y" +4y = 26 (1-/4): y(0) = 0, y(0) = 0 7. y" + 2y' + 2y = cost + 6(1 - m/2): y(0) = 0, y(0) = 0 8. y(4) – y = 6(t - 1): y(0) = 0, y'(0) = 0, y"(0) = 0, y"(0) = 0
Transcribed Image Text:Problems cach of Problems 1 through 8: A Find the solution of the given initial value problem. O b. Plot a graph of the solution. 1+2y' +2y = 6(1-); y(0) = 1, y'(0) = 0 2 y +4y = 6(t – ) - 6(1 – 27); y(0) = 0, y'(0) = 0. 3 Y+3y +2y = 6(1 - 5) + uj0(1): y(0) = 0, y'(0) = 1/2 4. y"+ 2y + 3y = sinr + 6(1 - 3m); y(0) = 0, y'(0) =0 5. y"+ y = 6(1- 2x) cos t; y(0) - 0, y(0) = 1 6. y" +4y = 26 (1-/4): y(0) = 0, y(0) = 0 7. y" + 2y' + 2y = cost + 6(1 - m/2): y(0) = 0, y(0) = 0 8. y(4) – y = 6(t - 1): y(0) = 0, y'(0) = 0, y"(0) = 0, y"(0) = 0
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