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s251ex1(fa02)

# s251ex1(fa02) - Math 251 Fall 2002 Midterm Exam 1 1(i...

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Math 251 Fall 2002 Midterm Exam 1 1. (i) Linear, homogeneous, oder 2. (ii) Nonlinear, order 1. (iii) Linear, nonhomogeneous, order 3. 2. (a) y 00 - 4 y 0 + 13 y = 0 λ 2 - 4 λ + 13 = 0 λ 1 , 2 = 2 ± 3 i y ( t ) = C 1 e 2 t cos 3 t + C 2 e 2 t sin 3 t (b) y 00 + 2 y 0 + y = 0 λ 2 + 2 λ + 1 = 0 λ 1 , 2 = - 1 y ( t ) = C 1 e - t + C 2 te - t 3. (a) y 0 = cos t + 1 y , y (0) = - 3 ydy = (cos t + 1) dt y 2 2 = sin t + t + const ( - 3) 2 2 = sin 0 + 0 + const const = 9 2 y ( t ) = + s 2 sin t + t + 9 2 (b) ty 0 + 3 y = 3 , y (1) = 2 y 0 + 3 t y = 3 t μ ( t ) e R 3 t dt = e 3 R = t 3 y ( t ) = R t 3 3 t dt t 3 = t 3 + const t 3 = 1 + const t 3 2 = 1 + const 1 3 const = 1 y ( t ) = 1 + 1 t 3 1

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4. (a) D (0) = 1000 , dD dt = 1 10 D ( t ) - 120 (b) D ( t ) = 120 1 10 + Ce 1 10 t 1000 = 1200 + Ce 0 C = - 200 D ( t ) = 1200 - 200 e 1 10 t (c) D ( t ) = 0 1200 - 200 e 1 10 t = 0 e 1 10 t = 6 t = 10 ln 6 5. (a) Both ln | t - 10 | and 1 t 2 - t must be continuous, that is t - 10 6 = 0 and t 2 - 2 6 = 0 and therefore we have 3 ”holes” in the real line: 0, 1, and 10. The maximal interval without holes, containing 8, is (1 , 10). (b) Both the function and its partial derivative with respect to y must be continuous: (1 - t 2 - y 2 ) 1 2 and 1 2 (1 - t 2 - y 2 ) - 1 2 ( - 2 y ) This means 1 - t 2 - y 2 > 0, that is t 2 0 + y 2 0 < 1 (point ( t 0 , y 0 ) must lie within the unit circle about the origin). 6. (a) The critical points are y = - 3 , 0 , 3. (b) The sign of the right-hand side of the equation changes according to the following pattern as y goes from -∞ to + : + - + - . Hence - 3 is stable, 0 is unstable, 3 is
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