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Dy Dx xy x 2 y 2

dy/dx=y^2/(xy-x^2),分子分母同时除以x^2 即dy/dx=(y/x)^2 / (y/x -1) 这时令y/x=u, 那么y=ux,所以dy/dx=u+x*du/dx, 代入得到u+x*du/dx= u^2/(u-1), 即x*du/dx=u/(u-1), 所以(1 -1/u)*du=1/x *dx, 两边积分得到u-lnu =lnx+C,(C为常数) ...

dy/dx = (x + y)² 令t = x + y,dt/dx = 1 + dy/dx dt/dx - 1 = t² dt/dx = (1 + t²) dt/(1 + t²) = dx arctan(t) = x + C₁ x + y = tan(x + C₁) y = tan(x + C₁) - x

请采纳,谢谢

如图中:

dy/dx=-x/y 即ydy=-xdx 两边积分 ∫ydy=∫-xdx 所以y²/2=(-x²+C)/2 y²=-x²+C 所以y=√(C-x²)

=x(x/y)^2十(y/x) 令y/x=t,y=tx dy=xdt十tdx 代入: xdt/dx十t=x/t^2十t dx=t^2dt x=t^3/3十C 回代: x=(y/x)^3/3十C

对函数两边求导,有: 2x+y+xy'-2yy'=0 2x+y+y'(x-2y)=0 y'=(2x+y)/(2y-x).

解:∵(3xy+x^2)dy+(y^2+xy)dx=0 ==>2y(3xy+x^2)dy+2y(y^2+xy)dx=0 (等式两端同乘2y) ==>2(3xy^2dy+y^3dx)+2(x^2ydy+xy^2dx)=0 ==>2d(xy^3)+d(x^2y^2)=0 ==>2∫d(xy^3)+∫d(x^2y^2)=0 ==>2xy^3+x^2y^2=C (C是常数) ∴此方程的通解是2xy^3+x^2y^2=C。

如图

I = ∮ (x²-xy³)dx+(y²-2xy)dy (格林公式) = ∫∫ (-2y+2xy)dxdy = ∫ dx ∫ 2(x-1)ydy = ∫ dx [(x-1)y^2] = ∫ 4(x-1)dx = [2x^2-4x] = 0

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