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Solve the equation x" +2x'+2x=25(t-π), x(0)=0, x'(0)=0 a. x(t)==—=sin(2t) x(t)=[1-e-²¹-2te-²¹] + µ(t− 2)(t − 2)e−2(t− 2t) x(t)=2μ(t-π)e-t-)sin(t-1) b. c. d. x(t)=[2-e²µ(t− π) + eªu(t−2π)]e-²tsint e. x(t) == [1 +µ(t - π)]sin(2t)

Solve the equation x" +2x'+2x=25(t-π), x(0)=0, x'(0)=0 a. x(t)==—=sin(2t) x(t)=[1-e-²¹-2te-²¹] + µ(t− 2)(t − 2)e−2(t− 2t) x(t)=2μ(t-π)e-t-)sin(t-1) b. c. d. x(t)=[2-e²µ(t− π) + eªu(t−2π)]e-²tsint e. x(t) == [1 +µ(t - π)]sin(2t)

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.CR: Chapter 9 Review
Problem 54CR
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Solve the equation x" + 2x' +2x = 25(t - π), x(0)=0, x'(0) = 0 = sin(2t) a. x(t)= b. x(t)= =[1-6 c. x(t)=2μ(t)e-(-) sin(t-1) d. x(t)=[2-e²u(t = π) +eªu(t−2π)]e-²¹ sint e. x(t) = O a O b O c Od O e -e-21-2te-21] + µ(t-2)(t-2)e-2 (1-21) = (1+µ +μ(t - π)]sin(2t)
Transcribed Image Text:Solve the equation x" + 2x' +2x = 25(t - π), x(0)=0, x'(0) = 0 = sin(2t) a. x(t)= b. x(t)= =[1-6 c. x(t)=2μ(t)e-(-) sin(t-1) d. x(t)=[2-e²u(t = π) +eªu(t−2π)]e-²¹ sint e. x(t) = O a O b O c Od O e -e-21-2te-21] + µ(t-2)(t-2)e-2 (1-21) = (1+µ +μ(t - π)]sin(2t)
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