如图所示：
let
u= y'
du/dy = d/dx(y' ) . (dx/dy)
= y''/y'
y''= u .du/dy
/
y''=e^(2y)
u .du/dy = e^(2y)
∫u du = ∫e^(2y) dy
(1/2)u^2 = (1/2)e^(2y) + C'
(y')^2 = e^(2y) + C
y'(0) = 1 , y(0)= 0
1^2 = e^0 + C
C = 0
ie
(y')^2 = e^(2y)
y' = e^y
∫dy/e^y = ∫dx
-e^(-y) = x + C1
y(0) =0
-1= 0+C1
=>
e^(-y) = x - 1
y = ln|1/(x-1)|
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024