求解一道高数题,谢谢!

2025-05-11 15:10:24
推荐回答(1个)
回答1:

let
u=y/x
du/dx = (1/x).dy/dx - (1/x^2)y
dy/dx =x[ du/dx + (1/x)u ]
/
y'=y/x + sec(y/x)
x[ du/dx + (1/x)u ] = u + secu
xdu/dx =secu
∫cosu du = ∫dx/x
sinu = ln|x| + C
sin(y/x) = ln|x| + C
y = xarcsin[ln|x| + C]