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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

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

dy=xydx 1/ydy=xdx ln|y|=x²/2+C ∴dy/dx=xy的通解为y=±e^(x²/2+C) e^(x²/2+C)表示±e的(x²/2+C)次方

解:令z=1/y,则y'=-z'/z^2 代入原方程,化简得 xz'-z=-axlnx ==>xdz-zdx=-axlnxdx ==>(xdz-zdx)/x^2=-alnxdx/x (等式两端同除x^2) ==>d(z/x)=-alnxd(lnx) ==>∫d(z/x)=-a∫lnxd(lnx) ==>z/x=C-a(lnx)^2/2 (C是常数) ==>1/(xy)=C-a(lnx)^2/2 ==>xy...

=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

答: x²+y²-xy=1 对x求导: 2x+2yy'-y-xy'=0 (2y-x)y'=y-2x y'=(y-2x) /(2y-x) 所以: dy / dx =(y-2x) /(2y-x)

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