方程两边对x求导得ye^(xy)+xy'e^(xy)+3y'y^2-5=0
所以dy/dx=y'=[5-ye^(xy)]/[3y^2+xe^(xy)],x=0时=(5-y)/(3y^2)
y''=(y')'={-[3y^2+xe^(xy)]e^(xy)[y^2+y'+xy'y]-[5-ye^(xy)][6y'y+(y'x^2+1+xy)e^(xy)]}/[3y^2+xe^(xy)]^2
={-[3y^2+xe^(xy)]e^(xy)[y^2+[5-ye^(xy)]/[3y^2+xe^(xy)]+xy[5-ye^(xy)]/[3y^2+xe^(xy)]]-[5-ye^(xy)][6y[5-ye^(xy)]/[3y^2+xe^(xy)]+(x^2[5-ye^(xy)]/[3y^2+xe^(xy)]+1+xy)e^(xy)]}/[3y^2+xe^(xy)]^2
=./[3y^2+xe^(xy)]^3
给点分吧..我快晕了