Question

求此题详解，多谢大佬！

Answer 1

y(0) =y'(0) =0
y''=e^(2y)
let
u=y'
du/dy = y''/y'
y''= u.du/dy
/
y''=e^(2y)
udu/dy = e^(2y)
∫u du = ∫ e^(2y) dy
(1/2)u^2 = (1/2) e^(2y) + C1
y(0) =y'(0) =0
0= (1/2) + C1
C1= -1/2
u^2 = e^(2y) -1
dy/dx =√[ e^(2y) -1 ]
∫dy/√[ e^(2y) -1 ] = ∫dx
arctan [√[ e^(2y) -1 ] ] = x+C2
y(0) =y'(0) =0
0=0+C2
=> C2=0
arctan [√[ e^(2y) -1 ] ] = x
√[ e^(2y) -1 ] = tanx
e^(2y)-1 = (tanx)^2
e^(2y) = (secx)^2
2y= 2ln|secx|
y =ln|secx|
/
let
e^y = secθ
e^y dy = secθ.tanθ dθ
dy = tanθ dθ
∫dy/√[ e^(2y) -1 ]
=∫tanθ dθ/ tanθ
=θ + C
=arctan [√[ e^(2y) -1 ] ] + C'