A5 Solutions

# 4 u v tan x 0 5 u v cot x sin x 6 dividing

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Unformatted text preview: ￿ sin x = 0 (3) −u￿ sin x + v ￿ cos x = sin2 x. (4) u￿ + v ￿ tan x = 0 (5) −u￿ + v ￿ cot x = sin x, (6) Dividing (3) by cos x and (4) by sin x yields and adding these yields v ￿ (tan x + cot x) = sin x. ￿ ￿ Multiplying through by tan x, we obtain v ￿ tan2 x + 1 = sin x tan x, i.e. 2 2 ￿ 2 ￿ v (sec x) = sin x sec x, so v = sin x cos x. Integrating, we have v= 13 sin x. 3 ￿ ￿ Now, from (5) we know that u￿ = −v ￿ tan x = − sin2 x cos x tan x = − sin3 x. Therefore ￿ u = − sin3 xdx AMATH 350 Assignment #5 Solutions - Fall 2011 Page 2 ￿ ￿ ￿ = − sin x sin2 x dx ￿ ￿ ￿ = − sin x 1 − cos2 x dx ￿ ￿ = − sin xdx + sin x cos2 xdx = cos x − 1 cos3 x. 3 We conclude that 1 1 cos4 x + sin4 x. 3 3 2 dy ￿2 ￿￿ ￿ 2d y + + 5y = 0, x &gt; x (1) 5.This can 5x simpliﬁed: sin4 0 −cos4 x = sin x + cos2 x sin2 x − cos2 x = x dx2 bedx dy d2 y Therefore 0 &gt; (1) 5.sin2 x2− cos2ln(.x),y x &gt; ,0.xApply chain rule: x2 Let z x x+ 5 = 0 (a) + 5 = yp = cos2 x...
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