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# practice_ch1 - EXERCISE 1.1 4 Second order nonlinear...

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EXERCISE 1.1 4. Second order; nonlinear because of cos( r + u ) 8. Second order; nonlinear because of x ˙ 2 10. Writing the differential equation in the form u ( dv/du ) + (1 + u ) v = ue u we see that it is linear in v . However, writing it in the form ( v + uv − ue u )( du/dv ) + u = 0, we see that it is nonlinear in u . 11. From y = e− x/ 2 we obtain y’ = 1/2 e −x/ 2 . Then 2 y’ + y = −e −x/ 2 + e −x/ 2 = 0. 14. From y = cos x ln(sec x + tan x ) we obtain y’ = 1 + sin x ln(sec x + tan x ) and y” = tan x + cos x ln(sec x + tan x ). Then y” + y = tan x . 18. The function is y = 1 /√ 1 sin x , whose domain is obtained from 1 sin x 0 or sin x 1. Thus, the domain is {x | x π/ 2 + 2 nπ} . From y’ = 1/2 (1 sin x ) 3 / 2 ( cos x ) we have 2 y’ = (1 sin x ) 3 / 2 cos x = [(1 sin x ) 1 / 2 ] 3 cos x = y 3 cos x. An interval of definition for the solution of the differential equation is ( π/ 2 , 5 π/ 2). Another one is (5 π/ 2 , 9 π/ 2), and so on. 21. Differentiating P = c 1 e t / (1 + c 1 e t ) we obtain 27. (a) From y = e mx we obtain y’ = me mx . Then y’ + 2 y

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practice_ch1 - EXERCISE 1.1 4 Second order nonlinear...

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