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(cos2ycosy-x)y'=ctgyy(1)=π/2,求微分方程的特解
题目详情
(cos2ycosy-x)y'=ctgy y(1)=π/2,求微分方程的特解
▼优质解答
答案和解析
(cos2ycosy-x)y'=coty
(cos2ysiny-xsiny/cosy]dy/dx=1
cos2ysinydy-xsinydy/cosy=dx
cos2ydcosy-xdcosy/cosy=dx
cos2ycosydcosy-xdcosy=cosydx
(cos2ycosy)dcosy=xdcosy+cosydy
(2cosy^3-cosy)dcosy=dxcosy
(1/2)[(cosy)^4-(cosy)^2]=xcosy+C
y(1)=π/2 C=0
特解 (1/2)[(cosy)^4-(cosy)^2]=xcosy
(cos2ysiny-xsiny/cosy]dy/dx=1
cos2ysinydy-xsinydy/cosy=dx
cos2ydcosy-xdcosy/cosy=dx
cos2ycosydcosy-xdcosy=cosydx
(cos2ycosy)dcosy=xdcosy+cosydy
(2cosy^3-cosy)dcosy=dxcosy
(1/2)[(cosy)^4-(cosy)^2]=xcosy+C
y(1)=π/2 C=0
特解 (1/2)[(cosy)^4-(cosy)^2]=xcosy
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