339_pdfsam_math 54 differential equation solutions odd

339_pdfsam_math 54 - 6 L min × 2 kg L = 1 2 kg min to the original y 12 kg min Thus the system(1 on page 242 becomes x = − 1 3 x 1 12 y 1 2 y =

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Review Problems y ( t )= C 3 e t + C 1 C 2 2 cos t + C 1 + C 2 2 sin t, z ( t )= C 3 e t C 1 C 2 2 sin t + C 1 + C 2 2 cos t. To fnd constants C 1 , C 2 ,and C 3 , we use the initial conditions. So we get 0= x (0) = C 1 cos 0 + C 2 sin 0 = C 1 , 0= y (0) = C 3 e 0 + C 1 C 2 2 cos 0 + C 1 + C 2 2 sin 0 = C 3 + C 1 C 2 2 , 2= z (0) = C 3 e 0 C 1 C 2 2 sin 0 + C 1 + C 2 2 cos 0 = C 3 + C 1 + C 2 2 , which simplifes to C 1 =0 , C 1 C 2 +2 C 3 =0 , C 1 + C 2 +2 C 3 =4 . Solving we obtain C 1 =0 , C 2 =2, C 3 =1andso x ( t )=2s in t, y ( t )= e t cos t +sin t, z ( t )= e t +cos t +sin t. 7. Let x ( t )and y ( t ) denote the mass oF salt in tanks A and B, respectively. The only di±erence
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Unformatted text preview: 6 L / min × . 2 kg / L = 1 . 2 kg / min to the original ( y/ 12) kg / min. Thus the system (1) on page 242 becomes x = − 1 3 x + 1 12 y + 1 . 2 , y = 1 3 x − 1 3 y. ²ollowing the solution in Section 5.1, we express x = 3 y + y From the second equation and substitute it into the frst equation. (3 y + y ) = − 1 3 (3 y + y ) + 1 12 y + 1 . 2 ⇒ 3 y + 2 y + 1 4 y = 1 . 2 . 335...
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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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