213_pdfsam_math 54 differential equation solutions odd

213_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.5 Thus a general solution on ( π/ (2 V ) ,π/ (2 V )) is y 1 ( t )= y h ( t )+ y p ( t C 1 cos t + C 2 sin t + 1 1 V 2 cos Vt. (4.3) Since y ( t ) 0for t ≤− (2 V ), the initial conditions for the above solution are y 1 ± π 2 V ² = y 0 1 ± π 2 V ² =0 . From (4.3) we obtain y 1 ± π 2 V ² = C 1 cos ± π 2 V ² + C 2 sin ± π 2 V ² y 0 1 ± π 2 V ² = C 1 sin ± π 2 V ² + C 2 cos ± π 2 V ² + V 1 V 2 . Solving the system yields C 1 = V V 2 1 sin π 2 V ,C 2 = V V 2 1 cos π 2 V , and y 1 ( t V V 2 1 sin π 2 V cos t + V V 2 1 cos π 2 V sin t + 1 1 V 2 cos Vt = V V 2 1 sin ± t + π 2 V ² 1 V 2 1 cos Vt, π 2 V <t< π 2 V . For t>π/ (2 V ) given equation is homogeneous, and its general solution,
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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