求「(3y-x)dx+(y-3x)dy」/(x+y)^3的原函数
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求「(3y-x)dx+(y-3x)dy」/(x+y)^3的原函数
∵令P(x,y)=(3y-x)/(x+y)^3,Q(x,y)=(y-3x)/(x+y)^3
则可求得αP/αy=αQ/αx
∴[(3y-x)dx+(y-3x)dy]/(x+y)^3存在原函数F(x,y)
∵由公式得
F(x,y)=∫P(x,0)dx+∫Q(x,y)dy
=∫(-1/x^2)dx+∫[(y-3x)/(x+y)^3]dy
=1/x-1+(x-y)/(x+y)^2-x/x^2+C1 (C1是常数)
=(x-y)/(x+y)^2+C1-1
=(x-y)/(x+y)^2+C (令C=C1-1)
∴所求原函数是(x-y)/(x+y)^2+C (C是常数).
则可求得αP/αy=αQ/αx
∴[(3y-x)dx+(y-3x)dy]/(x+y)^3存在原函数F(x,y)
∵由公式得
F(x,y)=∫P(x,0)dx+∫Q(x,y)dy
=∫(-1/x^2)dx+∫[(y-3x)/(x+y)^3]dy
=1/x-1+(x-y)/(x+y)^2-x/x^2+C1 (C1是常数)
=(x-y)/(x+y)^2+C1-1
=(x-y)/(x+y)^2+C (令C=C1-1)
∴所求原函数是(x-y)/(x+y)^2+C (C是常数).
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