174_pdfsam_math 54 differential equation solutions odd

# 174_pdfsam_math 54 differential equation solutions odd - (...

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Chapter 4 7. Solving the auxiliary equation, 2 r 2 +7 r 4=0 ,y ie lds r =1 / 2 , 4. Thus a general solution is given by u ( t )= c 1 e t/ 2 + c 2 e 4 t , where c 1 and c 2 are arbitrary constants. 9. The auxiliary equation for this problem is r 2 r 11 = 0, which has roots r = 1 ± 1+4 · 11 2 = 1 ± 3 5 2 . Thus, a general solution to the given equation is y ( t )= c 1 e (1+3 5) t/ 2 + c 2 e (1 3 5) t/ 2 . 11. Solving the auxiliary equation, 4 r 2 +20 r +25=(2 r +5) 2 = 0, we conclude that r = 5 / 2is its double root. Therefore, a general solution to the given diFerential equation is w ( t )= c 1 e 5 t/ 2 + c 2 te 5 t/ 2 . 13. The auxiliary equation for this problem is r 2 +2 r 8 = 0, which has roots r = 4 , 2. Thus, a general solution is given by y ( t )= c 1 e 4 t + c 2 e 2 t , where c 1 , c 2 are arbitrary constants. To satisfy the initial conditions, y (0) = 3, y 0 (0) =
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Unformatted text preview: ( t ) = − 4 c 1 e − 4 t + 2 c 2 e 2 t and solve the system y (0) = c 1 e − 4 · + c 2 e 2 · = c 1 + c 2 = 3 , y (0) = − 4 c 1 e − 4 · + 2 c 2 e 2 · = − 4 c 1 + 2 c 2 = − 12 ⇒ c 1 = 3 , c 2 = 0 . Therefore, the solution to the given initial value problem is y ( t ) = (3) e − 4 t + (0) e 2 t = 3 e − 4 t . 15. The auxiliary equation for this equation is r 2 + 2 r + 1 = ( r + 1) 2 = 0. We see that r = − 1 is a repeated root. Thus, two linearly independent solutions are y 1 ( t ) = e − t and y 2 ( t ) = te − t . This means that a general solution is given by y ( t ) = c 1 e − t + c 2 te − t . 170...
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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.

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