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339_pdfsam_math 54 differential equation solutions odd

# 339_pdfsam_math 54 differential equation solutions odd - 6...

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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 find 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 simplifies 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 = 1 and so x ( t ) = 2 sin 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 difference between this problem and the problem in Section 5.1 is that a brine solution ﬂows in tank A instead of pure water. This change affects the input rate for tank A only, adding
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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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