s251ex1(sp00)

# s251ex1(sp00) - u = 0. The system, therefore, has a natural...

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MATH 251 Exam 1 Feb 21, 2000 ANSWER KEY 1. (a) first order, non-linear, homogenerous, ordinary (b) second order, linear, non-homogeneous, ordinary (c) second order, non-linear, non-homogeneous, ordinary (d) second order, linear, homogeneous, ordinary 2. (a) 100 3 6 Q dt dQ - = , or 6 100 3 = + Q dt dQ ; Q (0) = 0, where Q ( t ) = the amount of dissolved salt in the tank at time t . (b) 100 / 3 200 200 ) ( t e t Q - - = (c) The limiting concentration is 2 100 200 100 ) ( lim = = t Q t . 3. ) 4 4 sin cos ( ) ( 2 2 - + + - = - π t t t t t y ; the largest interval is (0, ). 4. (a) 2 / 0 ) 3 2 ( 3 2 ) ( t t e y e t y + + - = - (b) y 0 = ±2/3; at this value y would approach 0 as t approaches . 5. (a) t t e C e C t y - + = 2 2 1 ) ( (b) t t e t C e C t y 2 2 2 1 ) ( - - + = (c) t e C t e C t y t t sin cos ) ( 2 2 2 1 - - + = 6. This (undamped) system is described by the equation 2 u ″ + 8

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Unformatted text preview: u = 0. The system, therefore, has a natural frequency of 2 / = = m k (radians/sec). Hence, resonance would occur if 2 = = . 7. (a) The 3 equilibrium solutions are: y = 0 (stable), y = T (unstable), and y = K (stable). (b) The population would die out if 0 &lt; y &lt; T , its size would become stable if y 0 T . The population would never increase without bound. 8. The complementary solution is t t c e t C e C t y 3 2 3 1 ) ( + = . Therefore, the particular solution should have the form: t A t A e t A A t A t A t Y t sin cos ) ( 6 5 3 2 4 3 2 2 1 + + + + + = ....
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## s251ex1(sp00) - u = 0. The system, therefore, has a natural...

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