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285lect29 - 1 Lecture 30 Example 1 x00 3x0 5x = 4 cos 5t...

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1 Lecture 30 Example 1 x 00 + 3 x 0 + 5 x = ° 4 cos 5 t: Complimentary equation: x 00 + 3 x 0 + 5 x = 0 : Characteristic equation: r 2 + 3 r + 5 = 0 Solution is: r = ° 3 2 + 1 2 i p 11 ; r = ° 3 2 ° 1 2 i p 11 : Transient solution: x tr ( t ) = e ° 3 2 t C 1 cos p 11 2 t + C 2 sin p 11 2 t: ! Steady periodic solution: x sp ( t ) = A cos 5 t + B sin 5 t ; x 0 sp ( t ) = ° 5 A sin 5 t + 5 B cos 5 t ; x 00 sp ( t ) = ° 25 ( A cos 5 t + B sin 5 t ) : x 00 + 3 x 0 + 5 x = ° 25 ( A cos 5 t + B sin 5 t ) + 3 ( ° 5 A sin 5 t + 5 B cos 5 t ) + 5 ( A cos 5 t + B sin 5 t ) = cos 5 t ( ° 25 A + 15 B + 5 A ) + sin t ( ° 25 B ° 15 A + 5 B ) = ° 4 cos 5 t ) ° 20 A + 15 B = ° 4 ° 15 A ° 20 B = 0 : ) ° 20 A + 15 B = ° 4 3 A + 4 B = 0 ) B = ° 3 4 A ) ° 20 A + 15 ° ° 3 4 ± A = ° 4 ) A = 16 125 ) B = ° 3 4 A = ° 3 4 16 125 = ° 12 125 : 1
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Steady periodic solution: x sp ( t ) = 16 125 cos 5 t ° 12 125 sin 5 t: s ° 16 125 ± 2 + ° 12 125 ± 2 = 4 25 : So, x sp ( t ) = 4 25 ° 25 4 16 125 cos 5 t ° 25 4 12 125 sin 5 t ± = 4 25 ° 4 5 cos 5 t ° 3 5 sin 5 t ± = 4 25 cos (5 t ° ° ) ; where cos ° = 4 5 and sin ° = ° 3 5 ) ° = ° 0 : 643 5 : Finally, x sp ( t ) ± 0 : 16 cos (5 t + 0 : 643 5) : General solution: x ( t ) = 0 : 16 cos (5 t + 0 : 643 5) | {z } Steady periodic solution + e ° 3 2 t
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