一道高数题求助

简单计算一下即可，答案如图所示

这题考察你对公式的记忆
根据导数公式有(tany)' = y'(secy)^2
带入得到(tany)' + x/(1+x^2) tany =x
令z=tany, 上式即z' +xz /(1+x^2)=x
然后直接带入y' +p(x)y =q(x)型的公式即可

换元法，再求解

let

u=tany
du/dx = (secy)^2 . y'
p(x) = x/(1+x^2)
e^[∫p(x) dx] =e^[ ∫x/(1+x^2) dx ] = e^[(1/2)ln(1+x^2)] =√(1+x^2)
''
y'(secy)^2 + [x/(1+x^2)]tany =x
du/dx +[x/(1+x^2)]u =x
√(1+x^2) .[du/dx +[x/(1+x^2)]u] =x.√(1+x^2)
d/dx ( u.√(1+x^2) ) =x.√(1+x^2)
u.√(1+x^2) =∫x.√(1+x^2) dx
=(1/2)∫√(1+x^2) d(1+x^2)
=(1/3)(1+x^2)^(3/2) +C
u= (1/3)(1+x^2) +C/√(1+x^2)
tany =(1/3)(1+x^2) +C/√(1+x^2)
y|x=0 =0
0=1/3 +C
C=-1/3
tany =(1/3)(1+x^2) -(1/3)[1/√(1+x^2)]
y=arctan { (1/3)(1+x^2) -(1/3)[1/√(1+x^2)] }