# If you have the seperable DE

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 wanna do the wrong thing

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James R
$$\frac{x}{2(x-2)(x-1)} dx= \frac{1}{y^2 -1}dy$$