x(1+y^2)+y(1+x^2)dy/dx=0 尤其是积分的步骤请详细讲解

dy/dx,y=(1+x+x^2)e^x dy/dx=x(1+y^2)/y通解 y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy dy/dx=(e^x+x)(1+y^2)通解 y=x^2+3x+1 求dy Y=x√(1-x^2 )+arcsinx,dy=? y=sin^2 {x/(x+1)} 求dy y=2^x-cosx/1-x 求dy x（1+x^2）dy=（y+x^2y-x^2）dx通解 求方程 dy/dx+x(y-x) +x^3(y-x)^2=1的通解~ [y+(x^2)y]dy=[x(y^2)-x]dx ,dy=x(e^x)(1+y^2)dx 这两题的通解 dx/dy=x/y+[cos（x/y）]∧2,y（0）=1 y(x + y + 1) dx + (x + 2y) dy = 0:运用正合方程式求解 解微分方程x^ydx=(1-y^2+x^2-x^2y^2)dy, y=e^2x+x^2-yarcsin(1／x),求y′及dy y=x^2x,求dy 求解微分方程dy/dx+x/2y=1/2 dy/dx=1+x+y^2+xy^2