xy-sin(πy^2)=0 求dy/dx
人气:161 ℃ 时间:2020-04-04 01:30:52
解答
y+xy'-cos(πy²)2πyy'=0
y=[2πycos(πy²)-x]y'
y'=y/[2πycos(πy²)-x]
即:dy/dx=y/[2πycos(πy²)-x]
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