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Exercise 6.1.2 Convert each second- or higher-order equation into an equivalent system of first-order ODEs, with initial conditions. Use x1,x2,. for the dependent variables of the new system, with x₁ = u, x2 = u', etc. (a) 3u" (t) +5u' (t) + 4u(t)=0 with u(0) = 7 and u' (0) = 5 (b) u"(t) +u' (t) +u(t)=0 with u(0) = 1 and u' (0) = 0 ... (c) 2u" (t) +2cos (u' (t))+u(t)=0 with u(0) = 3 and u() = -1 (d) u" (t)+u' (t)u(t) = 7 with u(1) = 2 and u' (1) = 4 (e) u'" (t) +2u"(t) +u' (t) +5u(t)=0 with u(0) = 1, u' (0) = 0 and u" (0) = -1. Hint: take x₁ = u, x2 = u', and x3 = u". Two of the ODEs are x₁ = x2 and x2 = x3. (f) u(4) (t) +u'" (t) +4u"(t)+5u' (t) +3u(t) = 0 with u(0) = 1, u' (0) = 0, u" (0) = -1 and u"" (0) = 4. Hint: take x₁ = u, x2 = u', x3 = u", and x4 = u"". Three of the ODEs are x1 = x2, x2 = x3, and x3 = x4. @

Exercise 6.1.2 Convert each second- or higher-order equation into an equivalent system of first-order ODEs, with initial conditions. Use x1,x2,. for the dependent variables of the new system, with x₁ = u, x2 = u', etc. (a) 3u" (t) +5u' (t) + 4u(t)=0 with u(0) = 7 and u' (0) = 5 (b) u"(t) +u' (t) +u(t)=0 with u(0) = 1 and u' (0) = 0 ... (c) 2u" (t) +2cos (u' (t))+u(t)=0 with u(0) = 3 and u() = -1 (d) u" (t)+u' (t)u(t) = 7 with u(1) = 2 and u' (1) = 4 (e) u'" (t) +2u"(t) +u' (t) +5u(t)=0 with u(0) = 1, u' (0) = 0 and u" (0) = -1. Hint: take x₁ = u, x2 = u', and x3 = u". Two of the ODEs are x₁ = x2 and x2 = x3. (f) u(4) (t) +u'" (t) +4u"(t)+5u' (t) +3u(t) = 0 with u(0) = 1, u' (0) = 0, u" (0) = -1 and u"" (0) = 4. Hint: take x₁ = u, x2 = u', x3 = u", and x4 = u"". Three of the ODEs are x1 = x2, x2 = x3, and x3 = x4. @

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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Exercise 6.1.2 Convert each second- or higher-order equation into an equivalent system of first-order ODEs, with initial conditions. Use x1,x2,. for the dependent variables of the new system, with x₁ = u, x2 = u', etc. (a) 3u" (t) +5u' (t) + 4u(t)=0 with u(0) = 7 and u' (0) = 5 (b) u"(t) +u' (t)+u(t)=0 with u(0) = 1 and u' (0) = 0 .... (c) 2u" (t) +2cos (u' (t))+u(t)=0 with u(0) = 3 and u() = -1 (d) u" (t)+u' (t)u(t) = 7 with u(1) = 2 and u' (1) = 4 (e) u'" (t) +2u"(t) +u' (t) +5u(t)=0 with u(0) = 1, u' (0) = 0 and u" (0) = -1. Hint: take x₁ = u, x2 = u', and x3 = u". Two of the ODEs are x₁ = x2 and x2 = x3. (f) u(4) (t) +u'" (t) +4u"(t)+5u' (t)+3u(t) = 0 with u(0) = 1, u' (0) = 0, u" (0) = -1 and u"" (0) = 4. Hint: take x₁ = u, x2 = u', x3 = u", and x4 = u"". Three of the ODEs are x1 = x2, x2 = x3, and x3 = x4. @
Transcribed Image Text:Exercise 6.1.2 Convert each second- or higher-order equation into an equivalent system of first-order ODEs, with initial conditions. Use x1,x2,. for the dependent variables of the new system, with x₁ = u, x2 = u', etc. (a) 3u" (t) +5u' (t) + 4u(t)=0 with u(0) = 7 and u' (0) = 5 (b) u"(t) +u' (t)+u(t)=0 with u(0) = 1 and u' (0) = 0 .... (c) 2u" (t) +2cos (u' (t))+u(t)=0 with u(0) = 3 and u() = -1 (d) u" (t)+u' (t)u(t) = 7 with u(1) = 2 and u' (1) = 4 (e) u'" (t) +2u"(t) +u' (t) +5u(t)=0 with u(0) = 1, u' (0) = 0 and u" (0) = -1. Hint: take x₁ = u, x2 = u', and x3 = u". Two of the ODEs are x₁ = x2 and x2 = x3. (f) u(4) (t) +u'" (t) +4u"(t)+5u' (t)+3u(t) = 0 with u(0) = 1, u' (0) = 0, u" (0) = -1 and u"" (0) = 4. Hint: take x₁ = u, x2 = u', x3 = u", and x4 = u"". Three of the ODEs are x1 = x2, x2 = x3, and x3 = x4. @
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