AMath350.F13.A5.Sol

# Dividing 3 by cos x and 4 by sin x yields 5 u0 v 0

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Unformatted text preview: Dividing (3) by cos x and (4) by sin x yields (5) u0 + v 0 tan x = 0 (6) u0 + v 0 cot x = sin x, and adding these yields v 0 (tan x + cot x) = sin x. Multiplying through by tan x, we obtain v 0 tan2 x + 1 = sin x tan x, i.e. v 0 (sec2 x) = sin2 x sec x, so v 0 = sin2 x cos x. Integrating, we have 13 sin x. 3 v= Now, from (5) we know that u0 = v 0 tan x = sin2 x cos x tan x = sin3 x. Therefore Z u= sin3 xdx Z = sin x sin2 x dx Z = sin x 1 cos2 x dx Z Z = sin xdx + sin x cos2 xdx = cos x 1 cos3 x. 3 We conclude that yp = cos2 x 1 1 cos4 x + sin4 x. 3 3 dy d2 y (1) 5.This can 5x simpliﬁed: sin4 0 cos4 x = sin2 x + cos2 x x2 2 + be + 5y = 0, x > x dx dx 2 2 sin x cos ln(.x), x > 0. Apply chain rule: (a) Let z = x Therefore sin2 x cos2 x = dy dz dy 1 = 1 dy 1 2 dy dy = yp = cos2 xdx dz cos2 dz + )sinx dx = dz . x x dx 3x 3 d dy d 1 dy 1 d dy 1 dy d2 y = = 1 dz =2x dx dz2 x2 dz 2 dx dx dx dx x 2 cos x + sin x = 3 1 d2 y 1 dy 1 dz d2 y 1 dy = = x dx dz 2 x2 dz 1 x2 dz 2 x2 dz = cos2 x + 1 becomes Using the bold face expressions, eq. (1)...
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## This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.

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