yˊ=1/xy*[sin(xy^2)]^2-y/2x答案是设u=xy^2 ,最后结果是uˊ=2/(sinu)^2
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yˊ=1/xy*[sin(xy^2)]^2-y/2x答案是设u=xy^2 ,最后结果是uˊ=2/(sinu)^2
yˊ=1/xy*[sin(xy^2)]^2-y/2x
答案是设u=xy^2 ,最后结果是uˊ=2/(sinu)^2
yˊ=1/xy*[sin(xy^2)]^2-y/2x答案是设u=xy^2 ,最后结果是uˊ=2/(sinu)^2
令u=xy²,则u’=y²+2xyy’……①(这个应该能明白吧)
把yˊ=1/[xy*sin²(xy²)]-y/2x带入①,得
uˊ=y²+2xy/xy*sin²(xy²)-y²
即uˊ=y²+2xy/xy*sin²(xy²)-y²
又∵xy²=u
∴u’=2/sin²u
u=∫2/sin²u du
u=-2ctanu+C
把u=xy²带入
xy²=-2ctan(xy²)+C
yˊ=1/xy*[sin(xy^2)]^2-y/2x答案是设u=xy^2 ,最后结果是uˊ=2/(sinu)^2
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