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Unformatted text preview: ' \$ Set 3: First Order ODEs - Part 2 Kyle A. Gallivan Department of Mathematics Florida State University Ordinary Differential Equations Fall 2009 & 1 % ' \$ Problem 3.1 Textbook, p. 39, Problem 1 y ′ + 3y = t + e−2t p(t) = 3, g (t) = t + e−2t if µ′ (t) = p(t)µ(t) = 3µ(t) then µ(t) = e3t and (µ(t)y )′ = µ(t)y ′ + µ′ (t)y = µ(t)g (t) = e3t (t + e−2t ) y (t) = Cµ−1 (t) + µ−1 (t) & 2 µ(t)g (t) dt % ' \$ Problem 3.1 Textbook, p. 39, Problem 1 y (t) = Cµ−1 (t) + µ−1 (t) e3t (t + e−2t ) dt y (t) = Ce−3t + e−3t = Ce−3t + e−3t et + te3t dt r s′ = rs − use integration by parts r′ s 1 3t with r = t r = 1, s = e , s = e 3 1 3t 1 3t 1 3t 1 3t 3t te dt = te − e dt = te − e 3 3 3 9 ′ & µ(t)g (t) dt ′ 3 3t % ' \$ Problem 3.1 Textbook, p. 39, Problem 1 1 3t 1 3t te dt = te − e 3 9 3t y (t) = Ce−3t + e−3t et + te3t dt 1 1 y (t) = Ce−3t + e−2t + t − 3 9 1 Exponentials go to 0 for any initial condition and y (t) → 1 t − 9 . 3 & 4 % ' \$ Problem 3.2 Textbook, p. 39, Problem 11 y ′ + y = 5(sin 2t) p(t) = 1, g (t) = 5(sin 2t) if µ′ (t) = p(t)µ(t) = µ(t) then µ(t) = et and (µ(t)y )′ = µ(t)y ′ + µ′ (t)y = µ(t)g (t) = et (5(sin 2t)) y (t) = Cµ−1 (t) + µ−1 (t) & 5 µ(t)g (t) dt % ' \$ Problem 3.2 Textbook, p. 39, Problem 11 y (t) = Cµ−1 (t) + µ−1 (t) y (t) = Ce−t + 5e−t recall µ(t)g (t) dt et (sin 2t) dt eau e (sin bu) du = 2 a(sin bu) − b(cos bu) 2 a +b au ∴ y (t) = Ce−t + e−t et (sin 2t) − 2et (cos 2t) = Ce−t + (sin 2t) − 2(cos 2t) The exponential goes to 0 for any initial condition and and y (t) → (sin 2t) − 2(cos 2t) as t → ∞. & 6 % ' \$ Problem 3.2 Textbook, p. 39, Problem 11 Note rather than use the general form of eau (sin bu) du one can derive the speciﬁc form for et (sin 2t) dt as follows: use integration by parts r s′ = rs − with r = sin 2t r ′ = 2 cos 2t, et (sin 2t) dt = et (sin 2t) − 2 & 7 r′ s s′ = et , s = et et (cos 2t) dt % ' \$ Problem 3.2 to get et (cos 2t) dt use integration by parts r s′ = rs − with r = cos 2t r ′ = −2 sin 2t, et (cos 2t) dt = et (cos 2t) + 2 r′ s s′ = et , s = et et (sin 2t) dt Note repetition of previous integral. & 8 % ' \$ Problem 3.2 Textbook, p. 39, Problem 11 Putting it all together yields: et (cos 2t) dt = et (cos 2t) + 2 et (sin 2t) dt et (sin 2t) dt = et (sin 2t) − 2 et (cos 2t) + 2 = et (sin 2t) − 2et (cos 2t) − 4 et (sin 2t) dt et (sin 2t) dt et (sin 2t) dt = et (sin 2t) − 2et (cos 2t) 5 1t e (sin 2t) dt = e (sin 2t) − 2et (cos 2t) 5 t as desired. & 9 % ' \$ Problem 3.3 Textbook, p. 39, Problem 12 2y ′ + y = 3t2 1 3 y ′ + y = t2 2 2 32 t 2 if µ′ (t) = p(t)µ(t) = µ(t) then µ(t) = e0.5t p(t) = 0.5, g (t) = 3 and (µ(t)y )′ = µ(t)y ′ + µ′ (t)y = µ(t)g (t) = e0.5t ( t2 ) 2 y (t) = Cµ−1 (t) + µ−1 (t) & 10 µ(t)g (t) dt % ' \$ Problem 3.3 Textbook, p. 39, Problem 12 y (t) = Cµ−1 (t) + µ−1 (t) y (t) = Ce−0.5t + e−0.5t − 0 .5 t = Ce 3 − 0 .5 t +e 2 use integration by parts µ(t)g (t) dt 3 e0.5t ( t2 ) dt 2 t2 e0.5t dt r s′ = rs − r′ s with r = t2 r ′ = 2t, s′ = e0.5t , s = 2e0.5t t2 e0.5t dt = 2t2 e0.5t − 4 & 11 te0.5t dt % ' \$ Problem 3.3 Textbook, p. 39, Problem 12 We have t2 e0.5t dt = 2t2 e0.5t − 4 te0.5t dt integrate by parts again r = t r ′ = 1, s′ = e0.5t , s = 2e0.5t te0.5t dt = 2te0.5t − 2 ∴ e0.5t = 2te0.5t − 4e0.5t t2 e0.5t dt = 2t2 e0.5t − 4 2te0.5t − 4e0.5t = 2t2 e0.5t − 8te0.5t + 16e0.5t & 12 % ' Problem 3.3 \$ Textbook, p. 39, Problem 12 t2 e0.5t dt = 2t2 e0.5t − 8te0.5t + 16e0.5t 3 y (t) = Ce−0.5t + e−0.5t t2 e0.5t dt 2 3 − 0 .5 t 2 0 .5 t − 0 .5 t 2t e − 8te0.5t + 16e0.5t ∴ y (t) = Ce +e 2 3 = Ce−0.5t + 2t2 − 8t + 16 = Ce−0.5t + 3t2 − 12t + 24 2 The exponential goes to 0 for any initial condition and and y (t) → 3t2 − 12t + 24 as t → ∞. & 13 % ' \$ Problem 3.4 Textbook, p. 39, Problem 4 y ′ + t−1 y = 3(cos 2t), t > 0 p(t) = t−1 , g (t) = 3(cos 2t) if µ′ (t) = p(t)µ(t) = t−1 µ(t) then 1 dt = ln|t|, t p(t)dt = R µ(t) = e p(t)dt = eln|t| = t and (µ(t)y )′ = µ(t)y ′ + µ′ (t)y = µ(t)g (t) = t(3(cos 2t)) y (t) = Cµ−1 (t) + µ−1 (t) & 14 µ(t)g (t) dt % ' \$ Problem 3.4 Textbook, p. 39, Problem 4 y (t) = Cµ−1 (t) + µ−1 (t) y (t) = Ct−1 + 3t−1 use integration by parts µ(t)g (t) dt t(cos 2t) dt r s′ = rs − with r = t r ′ = 1, s′ = (cos 2t), r′ s s = 0.5(sin 2t) t((cos 2t))dt = 0.5t(sin 2t) − 0.5 (sin 2t) = 0.5t(sin 2t) + 0.25(cos 2t) & 15 % ' \$ Problem 3.4 Textbook, p. 39, Problem 4 y (t) = Cµ−1 (t) + µ−1 (t) y (t) = Ct−1 + 3t−1 µ(t)g (t) dt t(cos 2t) dt y (t) = Ct−1 + 3t−1 0.5t(sin 2t) + 0.25(cos 2t) −1 = Ct 3 3 (cos 2t) + (sin 2t) + 2 4 t Ct−1 → 0 and cos 2t is bounded. ∴ y (t) → & 3 (sin 2t)as t → ∞ 2 16 % ' \$ Problem 3.5 Textbook, p. 39, Problem 8 (1 + t2 )y ′ + 4ty = (1 + t2 )−2 → y ′ + 4t(1 + t2 )−1 y = (1 + t2 )−3 p(t) = 4t(1 + t2 )−1 , g (t) = (1 + t2 )−3 if µ′ (t) = p(t)µ(t) = 4t(1 + t2 )−1 µ(t) then p(t) dt = 2 1 dz = 2 ln|1 + t2 | = 2 ln(1 + t2 ) z 2t dt = 2 2) (1 + t R µ(t) = e p(t)dt 2 ln(1+t2 ) =e = (1 + t2 )2 and (µ(t)y )′ = µ(t)y ′ + µ′ (t)y = µ(t)g (t) = & 17 1 (1 + t2 ) % ' \$ Problem 3.5 Textbook, p. 39, Problem 8 y (t) = Cµ−1 (t) + µ−1 (t) C 1 y (t) = + (1 + t2 )2 (1 + t2 )2 µ(t)g (t) dt 1 dt (1 + t2 ) tan−1 t C + = (1 + t2 )2 (1 + t2 )2 The ﬁrst term goes to 0 for any initial condition. Since |tan−1 t| ≤ π/2 the second term also goes to 0. & 18 % ' \$ Integrating Factor Summary y ′ + ay = 0 ⇓ y ′ + ay = g (t) ⇓ y ′ + p(t)y = g (t) ⇓ general solution:y (t) = 1 C+ µ(t) 1 y0 + IVP solution:y (t) = µ(t) & 19 µ(t)g (t) dt , µ(t) = p(t) dt t µ(s)g (s) ds , µ(t0 ) = 1 t0 % ...
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## This note was uploaded on 07/21/2011 for the course MAP 2203 taught by Professor Gallian during the Fall '09 term at FSU.

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