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m238sp05t2s - Dierential Equations test two solutions...

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Differential Equations test two solutions Monday, May 16, 2005 1 . For which values of t is the solution to the following differential equation guaranteed to exist? ( t 2 - 9) y + (cos t ) y + y = 2 t and y (0) = - 4 Rewrite the equation to isolate the highest derivative term: y + cos t t 2 - 9 y + 1 t 2 - 9 y = 2 t t 2 - 9 The coefficient functions are not continuous at t = ± 3. The existence & uniqueness theorem for second order linear equations states that a unique solution exists from the initial value of t = 0 up to values of t where the coefficient functions are discontinuous. So a solution will exist when - 3 < t < 3. 2 . Calculate the Wronskian of the functions f ( t ) = e 2 t and g ( t ) = e - t . W = f ( t ) g ( t ) f ( t ) g ( t ) = e 2 t e - t 2 e 2 t - e - t = - 3 e t 3 . Which of the following are linearly independent sets of functions? ( a ) { f ( t ) = t, g ( t ) = t 2 } ( b ) { f ( t ) = cos t, g ( t ) = t cos t } ( c ) { f ( t ) = e t - sin t, g ( t ) = sin t - e t } In (a) and (b), one function is not a constant multiple of the other, so each set is linearly independent. Things are not so wonderful in part (c), where the function g = ( - 1) f , so that this set of functions is not linearly independent.
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