简单分析一下，答案如图所示
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第1个回答 2015-03-08
y'= (1-y^2)tanx
∫dy/(1-y^2) = ∫tanx dx
(1/2)∫ ( 1/(1-y) + 1/(1+y) ) dy = ∫tanx dx
ln|(1+y)/(1-y)| = 2ln|cosx| + C
y(0) =2
ln3 = C
=>
ln|(1+y)/(1-y)| = 2ln|cosx| + ln3
(1+y)/(1-y) = 3(cosx)^2
1+y = 3(cosx)^2.(1-y)
[1+3(cosx)^2]y = 3(cosx)^2 -1
y =[3(cosx)^2 -1] / [1+3(cosx)^2]本回答被提问者采纳
∫dy/(1-y^2) = ∫tanx dx
(1/2)∫ ( 1/(1-y) + 1/(1+y) ) dy = ∫tanx dx
ln|(1+y)/(1-y)| = 2ln|cosx| + C
y(0) =2
ln3 = C
=>
ln|(1+y)/(1-y)| = 2ln|cosx| + ln3
(1+y)/(1-y) = 3(cosx)^2
1+y = 3(cosx)^2.(1-y)
[1+3(cosx)^2]y = 3(cosx)^2 -1
y =[3(cosx)^2 -1] / [1+3(cosx)^2]本回答被提问者采纳