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Unformatted text preview: Please notice that, the following solutions for final 2005 and 2006 are only for reference and may not be totally consistent with standard grading scheme. If you find any questions or mistakes, please feel free to contact me. Tutor: Liu Dongwen Email: ldxab@ust.hk Reference solutions of final 2005 1. (a) On separating variables, we obtain (3 + 2 y ) dy = 2cos2 xdx. Integrate on both sides, then we have Z y 2 (3 + 2 y ) dy = Z x 2cos2 xdx, 3 y + y 2 fl fl y 2 = sin2 x  x , y 2 + 3 y + 2 = sin2 x. If we solve above equation for y and make use of y (0) = 2, then y = 3 √ 1 + 4sin2 x 2 , ≤ x ≤ π 2 1 2 arcsin 1 4 . (b) From the result of (a), it is clear that y attains its minimum at sin2 x = 1, or at x = π 4 . And the minimum value of y is y min = 3 √ 5 2 . 2 2. (a) With μ ( t ) = exp Z t λdt = e λt , we have x ( t ) = e λt •Z t e λt ( a + be λt ) dt ‚ = a λ a λ e λt + bte λt . (b) We set ˙ x ( t ) = ae λt + be λt λbte λt = 0 , then t m = a + b λb . It is easy to see that ˙ x ( t ) > 0 when t < t m , and ˙ x ( t ) < 0 when t > t m , thus x ( t ) attains its maximum at t m . (c) Clearly, x ( t ) → a/λ as t → ∞ . 2 1 3. Denote x ( t ) the temperature of the coffee. Then with some constant k , one has ˙ x = k ( x 20) , x (0) = 90 , x (5) = 60 . Separating variables, it follows dx x 20 = kdt, Z x 90 dx x 20 = Z t dt, ln( x 20)  x 90 = kt, x ( t ) = 20 + 70 e kt . Applying x (5) = 60, we have k = 1 5 ln 4 7 and x ( t ) = 20 + 70 × 4 7 ¶ t/ 5 . 2 4. The characteristic equation r 2 + 2 r + 1 = 0 has a repeated root r = 1 . Thus the general solution is x ( t ) = ( c 1 + c 2 t ) e t , and we have ˙ x ( t ) = ( c 1 + c 2 c 2 t ) e t . Applying the initial conditions gives us x (0) = c 1 = 1 , ˙ x (0) = c 1 + c 2 = α. Then c 1 = 1 , c 2 = α + 1 and the initial value problem has the solution x ( t ) = [1 + ( α + 1) t ] e t . (b) Equivalently, we need to determine α such that 1+( α +1) t becomes negative for large t > . It is obvious that α < 1 is the answer. 2 5. As in the previous question, the general solution is x ( t ) = ( c 1 + c 2 t ) e t , and ˙ x ( t ) = ( c 1 + c 2 c 2 t ) e t . 2 The initial conditions gives that x (0) = c 1 = 0 , ˙ x (0) = c 1 + c 2 = u , or that c 1 = 0 , c 2 = u ....
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This note was uploaded on 01/28/2011 for the course MATH 150 taught by Professor T.qian during the Spring '09 term at HKUST.
 Spring '09
 T.Qian

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