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m245-sm2soln-f07

# m245-sm2soln-f07 - MATH 245 SAMPLE MIDTERM EXAM 2 SOLUTIONS...

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MATH 245 SAMPLE MIDTERM EXAM 2 SOLUTIONS Fall 2007 Prof. Alexander (1) Normalize: y 00 + 1 t y 0 - 4 t 2 y = 4 t , so g ( t ) = 4 t . Wronskian W ( t ) = ± ± ± ± t 2 t - 2 2 t - 2 t - 3 ± ± ± ± = - 4 t . Use variation of parameters: u 1 ( t ) = - Z g ( t ) y 2 ( t ) W ( t ) dt = - Z 4 t - 1 - 4 t - 1 dt = t, u 2 ( t ) = Z g ( t ) y 1 ( t ) W ( t ) dt = - Z 4 t 3 - 4 t - 1 dt = - t 5 5 . Particular solution Y ( t ) = u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ) = t 3 - t 3 5 = 4 5 t 3 , general solution y ( t ) = 4 5 t 3 + c 1 t 2 + c 2 t - 2 . (2)(a) Homogeneous ﬁrst: r 2 + 2 r = 0, roots r = 0 , - 2, general solution y ( t ) = c 1 + c 2 e - 2 t . Now use undetermined coeﬃcients for a particular solution: e - t is OK but a constant is not OK because it is a solution of the homogeneous equation. We must multiply it by t , yielding the form Y ( t ) = At + Be - t . Then Y 0 ( t ) = A - Be - t , Y 00 ( t ) = Be - t so we need 4 + e - t = Y 00 + 2 Y 0 = 2 A - Be - t . Thus

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m245-sm2soln-f07 - MATH 245 SAMPLE MIDTERM EXAM 2 SOLUTIONS...

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