180_pdfsam_math 54 differential equation solutions odd

180_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 Therefore, the auxiliary equation has a double root 2 and a root 2. The functions e 2 t , te 2 t ,and e 2 t form a linearly independent solution set. Therefore, a general solution in this problem is z ( t )= c 1 e 2 t + c 2 te 2 t + c 3 e 2 t . 45. By inspection, we see that r = 2 is a root of the auxiliary equation, r 3 +3 r 2 4 r 12 = 0. Dividing the polynomial r 3 r 2 4 r 12 by r 2 yields r 3 r 2 4 r 12 = ( r 2) ( r 2 +5 r +6 ) =( r 2)( r +2)( r +3) . Hence, two other roots of the auxiliary equation are r = 2and r = 3. The functions e 3 t , e 2 t e 2 t are three linearly independent solutions to the given equation, and a general solution is given by y ( t c 1 e 3 t + c 2 e 2 t + c 3 e 2 t . 47. First we ±nd a general solution to the equation y 0 y 0 = 0. Its characteristic equation, r 3 r = 0, has roots
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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 University of California, Berkeley.

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