MHB Solve by separation of variables

[find_the_fun]

Solve given differential equation by separation of variables

[math]\frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8}[/math]

So separate x and y terms

[math](xy-2x+4y-8) dy = (xy+3x-y-3)[/math] ugh I'm stuck:(

 

[MarkFL]

Re: solve by separation of variables

You want to factor the numerator and denominator of the right side, then you may separate variables.

 

[find_the_fun]

Re: solve by separation of variables

I can factor to [math]\frac{dy}{dx}=\frac{(x-1)(y+3)}{(x+4)(y-2)}[/math] and rewriting gives [math]\frac{(y-2)}{(y+3)} dy = \frac{(x-1)}{(x+4)} dx[/math]. Am I on the right track? I don't know how to integrate this.

 

[MarkFL]

Re: solve by separation of variables

Yes, you are correct. For the left side, consider:

$$\frac{y-2}{y+3}=\frac{y+3-5}{y+3}=1-\frac{5}{y+3}$$

Do the same kind of thing on the right side, and you should be able to integrate now.

 

[find_the_fun]

Re: solve by separation of variables

Yes, you are correct. For the left side, consider:

$$\frac{y-2}{y+3}=\frac{y+3-5}{y+3}=1-\frac{5}{y+3}$$

Do the same kind of thing on the right side, and you should be able to integrate now.

Is an alternative to [math]\frac{y+3-5}{y+3}[/math] doing polynomial division and seeing y+3 goes into y-2 once with a remainder of 5? I'm not super clear on the thought process of getting [math]1-\frac{5}{y+3}$$.

 

[MarkFL]

Re: solve by separation of variables

Is an alternative to [math]\frac{y+3-5}{y+3}[/math] doing polynomial division and seeing y+3 goes into y-2 once with a remainder of 5? I'm not super clear on the thought process of getting [math]1-\frac{5}{y+3}[/math].

Yes, although the remainder is actually -5, but then you get the same result. I just find it simpler to do as I did above. To make what I did more clear, consider:

$$\frac{y-2}{y+3}=\frac{(y+3)+(-2-3)}{y+3}=\frac{y+3}{y+3}-\frac{5}{y+3}=1-\frac{5}{y+3}$$