求式子的微分方程满足所给初始条件的特解：cosydx+(1+e^-x)sinydy=0,yx=0=π/4;yx之间有一竖,x=0显示的小

Question

求式子的微分方程满足所给初始条件的特解：cosydx+(1+e^-x)sinydy=0,yx=0=π/4;yx之间有一竖,x=0显示的小

Answer 1

cosydx+(1+e^-x)sinydy=0
e^xcosydx+(e^x+1)sinydy=0
e^xcosydx+e^xsinydy=-sinydy
cosyde^x-e^xdcosy=-sinydy
de^x/cosy-e^xdcosy/cosy^2=-sinydy/cosy^2
d(e^x/cosy)=-d(1/cosy)
通解e^x/cosy=-1/cosy+C
即e^x=-1+Ccosy
y|x=0 =π/4
1=-1+√2C/2
√2C=4
C=2√2
特解e^x=-1+2√2cosy