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Exercise 2.1.1 Find a general solution to each linear ODE, and then find the specific solution with the given initial condition by using the integrating factor technique from Section 2.1.2. (a) u' (t) = u(t) +3, u(0) = = 3 (b) u' (t) = 2u(t) +4, u(0) = 0 (c) u' (t) = -3u(t) +3, u(0) = 5 (d) u' (t) = -3u(t) +9t, u(0) = 5 (e) u' (t) = u(t) + 2 sin(t), u(0) = 1 (f) u' (t) = -4u(t) +e', u(0) = 2 (g) u' (t) = tu(t) +t, u(0) = 2 (h) u' (t) = u(t)/t+2, u(1) = 3 (i) u' (t) = sin(t)u(t) + sin(t), u(0) = 4 (j) u' (t) = au(t) + b, u(0) = uo, where a, b, and uo are constants

Exercise 2.1.1 Find a general solution to each linear ODE, and then find the specific solution with the given initial condition by using the integrating factor technique from Section 2.1.2. (a) u' (t) = u(t) +3, u(0) = = 3 (b) u' (t) = 2u(t) +4, u(0) = 0 (c) u' (t) = -3u(t) +3, u(0) = 5 (d) u' (t) = -3u(t) +9t, u(0) = 5 (e) u' (t) = u(t) + 2 sin(t), u(0) = 1 (f) u' (t) = -4u(t) +e', u(0) = 2 (g) u' (t) = tu(t) +t, u(0) = 2 (h) u' (t) = u(t)/t+2, u(1) = 3 (i) u' (t) = sin(t)u(t) + sin(t), u(0) = 4 (j) u' (t) = au(t) + b, u(0) = uo, where a, b, and uo are constants

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.6: Variation
Problem 2E
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Exercise 2.1.1 Find a general solution to each linear ODE, and then find the specific solution with the given initial condition by using the integrating factor technique from Section 2.1.2. (a) u' (t) = u(t) +3, u(0) = 3 (b) u' (t) = 2u(t) +4, u(0) = 0 (c) u' (t) = −3u(t) +3, u(0) = 5 (d) u' (t) = -3u(t) +9t, u(0) = 5 (e) u' (t) = u(t) + 2 sin(t), u(0) = 1 (f) u'(t) = −4u(t) + e¹, u(0) = 2 (g) u' (t) = tu(t) +t, u(0) = 2 (h) u' (t) = u(t)/t+2, u(1) = 3 (i) u' (t) = sin(t)u(t) + sin(t), u(0) = 4 (j) u'(t) = au(t) + b, u(0) = uo, where a, b, and uo are constants
Transcribed Image Text:Exercise 2.1.1 Find a general solution to each linear ODE, and then find the specific solution with the given initial condition by using the integrating factor technique from Section 2.1.2. (a) u' (t) = u(t) +3, u(0) = 3 (b) u' (t) = 2u(t) +4, u(0) = 0 (c) u' (t) = −3u(t) +3, u(0) = 5 (d) u' (t) = -3u(t) +9t, u(0) = 5 (e) u' (t) = u(t) + 2 sin(t), u(0) = 1 (f) u'(t) = −4u(t) + e¹, u(0) = 2 (g) u' (t) = tu(t) +t, u(0) = 2 (h) u' (t) = u(t)/t+2, u(1) = 3 (i) u' (t) = sin(t)u(t) + sin(t), u(0) = 4 (j) u'(t) = au(t) + b, u(0) = uo, where a, b, and uo are constants
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