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Unformatted text preview: 1 ( x,t ) is indeed a solution. Similar procedure for u 2 ( x,t ) shows that it is also a solution (2.1.33) (6 pts ) The given ODE is rst order and linear. y + ay = be-t The integrating factor is = e R a dt = e at Therefore ( ye at ) = e at ( be-t ) = ye at = Z e at ( be-t ) dt = b Z e ( a- ) t dt If a = , then we have ye at = b Z 1 dt = bt + C = y = bte-at + Ce-at Since both exponentials are negative ( a > 0), y 0 as t . ( evaluate the limit as t ) If a 6 = ye at = b 1 a- e ( a- ) t + C 2 = y = b 1 a- e (- ) t + Ce-at Again, both exponentials are negative ( a > 0 and > 0 ), hence by similar reasoning, y 0 as t ) 3...
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