final07 - Final 2007 reference solution 1(a This is a...

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Final 2007 reference solution 1. (a) This is a separable equation: e - 2 y dy = - e - 2 x dx, - 1 2 e - 2 y y 0 = 1 2 e - 2 x x 0 , y = - 1 2 log(2 - e - 2 x ) . (b) From (a) we see that y exists if and only if 2 - e - 2 x > 0 , i.e. x > - (log 2) / 2 . 2. (a) The first order equation can be rewritten as y 0 + (1 + 1 x ) y = e - x , thus we get μ ( x ) = exp Z x 1 (1 + 1 x ) dx = xe x - 1 , y ( x ) = x - 1 e 1 - x Z x 1 ξe ξ - 1 e - ξ = 1 2 e - x ( x - x - 1 ) . (b) Clearly y ( x ) 0 as x → ∞ , thus the maximum will be obtained at some point x where y 0 ( x ) = 0 , i.e. y 0 ( x ) = 1 2 e - x (1 + x - 2 - x + x - 1 ) = 0 , from which we derive a cubic equation x 3 - x 2 - x - 1 = 0 . 3. By the hint one has F = mv dv dx = - kv, mdv = - kdx, Z 0 v 0 mdv = - k Z x 0 dx, x = mv 0 k . 4. (a) The characteristic equation is r 2 + 4 r + 5 = 0 , which has two roots r = - 2 ± i, thus the general solution is x = e - 2 t ( A cos t + B sin t ) . Apply the conditions x (0) = 2, ˙ x (0) = - 5, we get A = 2, B = - 1
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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.

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final07 - Final 2007 reference solution 1(a This is a...

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