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検索キーワード「homogeneous differential equation」に一致する投稿を表示しています

[最も選択された] dy/dx=x^2 y^2 1/2xy 330782-15. (x^(2)+y^(2)+2xy+1)(dy)/(dx)=x+y

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The equation is not exact, because you'd actually need d (2xy)/dy=d (x 2 y 2 )/dx, which means that 2x=2x Letting (x 2 y 2 )=M, 2xy=N, then Mx=2x, Ny=2x So MxNy=4x Now, (MxNy)/N=4x/ (2xy)=2/y, so this is just a function of y v (y) That means that we can let u (y)=e integral of vdy u (y)=e 2ln y =1/y 2