Quiz5Asolutions

Quiz5Asolutions - y π 4 = e-π 4 c 1 cos π 4 c 2 sin π 4 = √ 2 2 e-π 4 c 1 c 2 = 2 y π 4 = e-π 4 c 1-cos π 4-sin π 4 c 2-sin π 4 cos π

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MATH 415.01 QUIZ 5A Feb 10, 2010 SHOW ALL YOUR WORK! GOOD LUCK! NAME: 1. (6 pts.) Find the Wronskian of two solutions of the given differential equation below without solving the equation. t 2 y 00 - t ( t + 2) y 0 + ( t + 2) y = 0 Rewrite the equation: y 00 - t +2 t y 0 + t +2 t 2 y = 0 W = ce - R - t +2 t dt = ce R t +2 t dt = ce R (1+ 2 t ) dt = ce t +2 ln ( t ) = ct 2 e t 2. (6.5 pts.) Find the solution of the given initial value problem below. y 00 + 2 y 0 + 2 y = 0 , y ( π/ 4) = 2 , y 0 ( π/ 4) = - 2 Characteristic Equation: r 2 + 2 r + 2 = 0 . roots are - 1 + i and - 1 - i . Then, the solution is: y ( t ) = c 1 e - t cos t + c 2 e - t sin t Also, we have y 0 ( t ) = c 1 ( - e - t cos t - e - t sin t ) + c 2 ( - e - t sin t + e - t cos t ) Use initial conditions:
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Unformatted text preview: y ( π/ 4) = e-π/ 4 ( c 1 cos π/ 4 + c 2 sin π/ 4) = √ 2 / 2 e-π/ 4 ( c 1 + c 2 ) = 2 y ( π/ 4) = e-π/ 4 ( c 1 (-cos π/ 4-sin π/ 4) + c 2 (-sin π/ 4 + cos π/ 4)) = e-π/ 4 c 1 (-√ 2) =-2 The last equation implies that c 1 = √ 2 e π/ 4 and then using the first equation we get c 2 = √ 2 e π/ 4 The final solution is y ( t ) = √ 2 e π/ 4-t cos t + √ 2 e π/ 4-t sin t...
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This note was uploaded on 07/26/2011 for the course MATH 415 taught by Professor Costin during the Spring '07 term at Ohio State.

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