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Use the method of variation of parameters to find a particular solution of the follow- ing equation. Note that this equation is the same as Problem 4(a), but now you are using a different method to solve it. y" — 4y' + 4y = e²t We should end up with two equations - u'₁(t)y¹(t) +u2(t)y2(t) = g(t). u₁(t)y1(t) +u2(t)y2(t) = 0, We first note the two solutions for the homogeneous equation: Y₁ = e²t, Y2 = te²t From the above two equations u₁+u₁t = 0 2u'₁ +u'½(2t + 1) = 1. Plugging the result of the first equation into the second one: Namely, and Therefore, −2ut + u₂ (2t + 1) = 1 u₁ = 1 u₂ = t u'₁ ==t 1 azi m -}pe 1 1 Уз *3 = − Pe² + Pe² = 2² e²+

Use the method of variation of parameters to find a particular solution of the follow- ing equation. Note that this equation is the same as Problem 4(a), but now you are using a different method to solve it. y" — 4y' + 4y = e²t We should end up with two equations - u'₁(t)y¹(t) +u2(t)y2(t) = g(t). u₁(t)y1(t) +u2(t)y2(t) = 0, We first note the two solutions for the homogeneous equation: Y₁ = e²t, Y2 = te²t From the above two equations u₁+u₁t = 0 2u'₁ +u'½(2t + 1) = 1. Plugging the result of the first equation into the second one: Namely, and Therefore, −2ut + u₂ (2t + 1) = 1 u₁ = 1 u₂ = t u'₁ ==t 1 azi m -}pe 1 1 Уз *3 = − Pe² + Pe² = 2² e²+

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Use the method of variation of parameters to find a particular solution of the follow- ing equation. Note that this equation is the same as Problem 4(a), but now you are using a different method to solve it. y" — 4y' + 4y = e²t We should end up with two equations - u'₁(t)y¹(t) +u2(t)y2(t) = g(t). u₁(t)y1(t) +u2(t)y2(t) = 0, We first note the two solutions for the homogeneous equation: Y₁ = e²t, Y2 = te²t From the above two equations u₁+u₁t = 0 2u'₁ +u'½(2t + 1) = 1. Plugging the result of the first equation into the second one: Namely, and Therefore, −2ut + u₂ (2t + 1) = 1 u₁ = 1 u₂ = t u'₁ ==t 1 azi m -}pe 1 1 Уз *3 = − Pe² + Pe² = 2² e²+
Transcribed Image Text:Use the method of variation of parameters to find a particular solution of the follow- ing equation. Note that this equation is the same as Problem 4(a), but now you are using a different method to solve it. y" — 4y' + 4y = e²t We should end up with two equations - u'₁(t)y¹(t) +u2(t)y2(t) = g(t). u₁(t)y1(t) +u2(t)y2(t) = 0, We first note the two solutions for the homogeneous equation: Y₁ = e²t, Y2 = te²t From the above two equations u₁+u₁t = 0 2u'₁ +u'½(2t + 1) = 1. Plugging the result of the first equation into the second one: Namely, and Therefore, −2ut + u₂ (2t + 1) = 1 u₁ = 1 u₂ = t u'₁ ==t 1 azi m -}pe 1 1 Уз *3 = − Pe² + Pe² = 2² e²+
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