Solve the differential equation:

`dy/dx` = (2x - 6) /(x^2 - 2x)

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Solve `(dy)/(dx)=((2x-6))/((x^2-2x))`

`int dy = int ((2x-6))/((x^2-2x))dx`

`y=int ((2x-6))/((x^2-2x))`

Use the method of partial fractions to rewrite the integrand:

`(2x-6)/(x^2-2x)=(2x-6)/(x(x-2))=A/x+B/(x-2)`

So `A(x-2)+B(x)=2x-6==>A+B=2,-2A=-6`

Then A=3 and B=-1

`y=int(3/x-1/(x-2))dx`

`=3int1/x dx - int 1/(x-2)dx`

`=3ln|x|-ln|x-2|+C`

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We can check the solution. ```d/(dx)[3ln|x|-ln|x-2|+C]=3/x-1/(x-2)`

`=(3(x-2)-x)/(x(x-2))=(2x-6)/(x^2-2x)` as required.

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