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wksht05sol - MATH 152 L1 L2 Applied Linear Algebra and...

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MATH 152 L1 & L2 Applied Linear Algebra and Differential Equations Spring 2004-05 ¥ Week 05 Worksheet Solutions : Second-Order Linear Equations Q1. (The method of variation of parameters) The homogeneous equation ty 00 - ( t + 1) y 0 + y = 0 ( t > 0) has two linearly independent solutions y 1 ( t ) = t + 1 and y 2 ( t ) = e t . By the method of variation of parameters, the nonhomogeneous equation ty 00 - ( t + 1) y 0 + y = t 2 , t > 0 has solutions of the form u 1 ( t ) · ( t + 1) + u 2 ( t ) · e t for suitable choices of functions u 1 ( t ) and u 2 ( t ). (a) Assuming that u 0 1 ( t ) · ( t +1)+ u 0 2 ( t ) · e t = 0, derive the additional condition (another equation) that u 0 1 ( t ) and u 0 2 ( t ) must satisfy for u 1 ( t ) · ( t + 1) + u 2 ( t ) · e t to be a solution of the nonhomogeneous equation. Solution Since the nonhomogeneous equation has solutions of the form y ( t ) = u 1 ( t ) · ( t + 1) + u 2 ( t ) · e t , (1) then by differentiation, we have y 0 ( t ) = u 0 1 ( t ) · ( t + 1) + u 0 2 ( t ) · e t + u 1 ( t ) + u 2 ( t ) · e t . Let u 0 1 ( t ) · ( t + 1) + u 0 2 ( t ) · e t = 0. Then we have y 0 ( t ) = u 1 ( t ) + u 2 ( t ) · e t , (2) y 00 ( t ) = u 0 1 ( t ) + u 0 2 ( t ) · e t + u 2 ( t ) · e t . (3) Substituting (1)–(3) into the nonhomogeneous equation, we have t £ u 0 1 ( t ) + u 0 2 ( t ) · e t + u 2 ( t ) · e t / - ( t + 1) £ u 1 ( t ) + u 2 ( t ) · e t / + £ u 1 ( t ) · ( t + 1) + u 2 ( t ) · e t / = t 2 , or upon simplifications ( t 6 = 0), we have u 0 1 ( t ) + u 0 2 ( t ) · e t = t .
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