xy-sin(πy^2)=0 求dy/dx
来源:学生学帮网 编辑:学帮网 时间:2024/06/25 06:30:20
xy-sin(πy^2)=0 求dy/dx
y+xy'-cos(πy²)2πyy'=0
y=[2πycos(πy²)-x]y'
y'=y/[2πycos(πy²)-x]
即:dy/dx=y/[2πycos(πy²)-x]
两边对x求导得
y+xy'-cos(πy^2)*(πy^2)'=0
y+xy'-πcos(πy^2)*2yy'=0
xy-sin(πy^2)=0 求dy/dx
y+xy'-cos(πy²)2πyy'=0
y=[2πycos(πy²)-x]y'
y'=y/[2πycos(πy²)-x]
即:dy/dx=y/[2πycos(πy²)-x]
两边对x求导得
y+xy'-cos(πy^2)*(πy^2)'=0
y+xy'-πcos(πy^2)*2yy'=0