有没有会的,帮忙做一下,过程详细点,谢谢

2025-05-11 00:55:26
推荐回答(1个)
回答1:

y'+ycosx = e^(-sinx)
let
u= e^(sinx) .y
du/dx = e^(sinx) . dy/dx + ycosx . e^(sinx )
y'+ycosx = e^(-sinx)

e^(sinx). y' + ycosx . e^(sinx ) =1
du/dx =1
u = x +C
e^(sinx) .y = x+C
y = (x+C) .e^(-sinx)