172_pdfsam_math 54 differential equation solutions odd

172_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 4 Writing y (t) in the form √ 1 2 5 √ sin 3t + √ cos 3t = 5 sin(3t + γ ), 5 5 √ √ where γ = arctan(1/2), we conclude that |y (t)| = 5| sin(3t + γ )|, and so max |y (t)| = 5 y (t) = (since max | sin(3t + γ )| = 1). 5. We differentiate y (t) twice and obtain √ y (t) = e−2t sin( 2t) √ √ √ y (t) = e−2t [(−2) sin( 2t) + 2 cos( 2t)] √ √ √ √ √ √ √ y (t) = e−2t (−2)2 sin( 2t) + (−2) 2 cos( 2t) + (−2) 2 cos( 2t) − ( 2)2 sin( 2t) √ √ √ = e−2t 2 sin( 2t) − 4 2 cos( 2t) . Substituting these functions into the differential equation, we get √ √ √ my + by + ky = y + 4y + 6y = e−2t 2 sin( 2t) − 4 2 cos( 2t) √ √ √ √ +4e−2t [(−2) sin( 2t) + 2 cos( 2t)] + 6e−2t sin( 2t) √ √ √ √ = e−2t (2 − 8 + 6) sin( 2t) + (−4 2 + 4 2) cos( 2t) = 0. √ √ Therefore, y = e−2t sin( 2t) is a solution. As t → +∞, e−2t → 0 while sin( 2t) remains bounded. Therefore, lim y (t) = 0. t→+∞ √ 7. For y = A cos 5t + B sin 5t, y = −5A sin 5t + 5B cos 5t, y = −25A cos 5t − 25B sin 5t. Inserting y , y , and y into the given equation and matching coefficients yield y + 2y + 4y = 3 sin 5t ⇒ (−25A cos 5t − 25B sin 5t) + 2(−5A sin 5t + 5B cos 5t) + 4(A cos 5t + B sin 5t) = (−21A + 10B ) cos 5t + (−10A − 21B ) sin 5t = 3 sin 5t ⇒ −21A + 10B = 0, −10A − 21B = 3 ⇒ A = −30/541, B = −63/541. Thus, y = −(30/541) cos 5t−(63/541) sin 5t is a synchronous solution to y + 2y + 4y = 3 sin 5t. 168 ...
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