Question

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​

Answer 1

It looks like the differential equation is

`(dy)/(dx) = (xy + 3x - y - 3)/(xy - 2x + 4y - 8)`

Factorize the right side by grouping.

`xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)`

`xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)`

Now we can separate variables as

`(dy)/(dx) = ((x-1)(y+3))/((x+4)(y-2)) \implies (y-2)/(y+3) \, dy = (x-1)/(x+4) \, dx`

Integrate both sides.

`\displaystyle \int (y-2)/(y+3) \, dy = \int (x-1)/(x+4) \, dx`

`\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx`

`\implies \boxed = x - 5 \ln`

You could go on to solve for
`y` explicitly as a function of
`x`, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.