- #1
schattenjaeger
- 178
- 0
If you have the seperable DE...
dy/dx=[x(y^2-2)]/(2x^2-6x+4)
that eventually ends up
(xdx)/(2x^2-6x+4)=dy/(y^2-1), right
'cuz that's some integration I REALLY don't feel like doing by hand, so I don't want to do the wrong thing
dy/dx=[x(y^2-2)]/(2x^2-6x+4)
that eventually ends up
(xdx)/(2x^2-6x+4)=dy/(y^2-1), right
'cuz that's some integration I REALLY don't feel like doing by hand, so I don't want to do the wrong thing