Unformatted text preview: , − D [ x ] + y ( t ) + ( D 2 + D + 1) [ y ] = 0 . By multiplying the ±rst equation above by D and the second by ( D 2 + D + 1) and adding the resulting equations, we can eliminate the function y ( t ). Thus, we have n ( D 2 + D + 1 ) 2 − D 2 o [ x ] = 0 ⇒ ±²( D 2 + D + 1 ) − D ³ · ²( D 2 + D + 1 ) + D ³´ [ x ] = 0 ⇒ ±( D 2 + 1 ) ( D + 1) 2 ´ [ x ] = 0 . This last equation is a fourth order linear diFerential equation with constant coeﬃcients whose associated auxiliary equation has roots r = − 1, − 1, i , and − i . Therefore, the solution to this diFerential equation is x ( t ) = c 1 e − t + c 2 te − t + c 3 cos t + c 4 sin t 313...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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