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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online