Solve the ordinary differential equation. 
Solve the equation: $$x(2y+1)dx=y(x^2-3x+2)dy$$

To solve this, I first move $y(x^2-3x+2)dy$ to the LHS, then divide both sides by $(2 y + 1)(x^2 - 3 x + 2)$, the equation becomes:
$$\dfrac{x}{x^2-3x+2}dx-\dfrac{y}{2y+1}dy=0$$
After that, I integrate both sides: $$\int\dfrac{x}{x^2-3x+2}dx-\int\dfrac{y}{2y+1}dy=C$$
Which gives: $$2\ln|x-2|-\ln|x-1|+\dfrac{\ln|2y+1|-2y-1}{4}=C$$
Did I do this correctly?

You want to get all the y stuff to one side, and all the x stuff to the other. These are called "separable equations" for this reason.
$x(2y+1)dx=y(x^2-3x+2)dy$ $\Rightarrow$
$\frac{x}{x^2-3x+2}dx=\frac{y}{2y-1}dy$ $\Rightarrow$
$\int\frac{x}{x^2-3x+2}dx=\int\frac{y}{2y-1}dy$
Now we integrate.
The left side looks like partial fraction decomposition and the right side looks like a u-sub with $u=2y-1$.
