given that y=(1+4x)e^-2x show that d^2y/dx^2+2dy/dx+xy=0

given that y=(1+4x)e^-2x show that d^2y/dx^2+2dy/dx+xy=0 d^2y /dx^2 - 24 =0 dy/dx -2y = e^-x dy/dx,y=(1+x+x^2)e^x dy/dx+(e^((y^2)+x))/y=0 dy/dx=1+x+y^2+xy^2 dy/dx=(x^4+y^3)/xy^2 x^2+xy+y^3=1,求dy/dx dy/dx=(e^x+x)(1+y^2)通解 y=f（x）由方程xy+e^xy+y=e确定,求dy/dx和d^2y/dx^2 e^xy+x+y=2求dy/dx |x=1 由2xy+e∧x-e∧y=0求dy/dx 已知e^y+e^x-xy^2=0,求dy/dx