# Thread: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

2. ## Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

Two minus signs dropped when you rationalise the fractions in line three.

3. ## Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0 Originally Posted by Archie Two minus signs dropped when you rationalise the fractions in line three.
i still cant find the mistake ? what is the correct one ?

4. ## Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0 Originally Posted by Archie Two minus signs dropped when you rationalise the fractions in line three.
well , i have redo the question , yet i still cant get (x+1)^2 and (y-1)^2 ....... 5. ## Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

You just need to add 2 two both sides. Then the right hand side $c+2 = c_1$ is still an arbitrary constant.

6. ## Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0 Originally Posted by Archie You just need to add 2 two both sides. Then the right hand side $c+2 = c_1$ is still an arbitrary constant.

why the author wanna do so ?

7. ## Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0

So it looks nicer. Or perhaps he integrated $\displaystyle (x-1)$ as a single term. You could even do substitutions before you start.

8. ## Re: (x^2)(y+1)dx + (y^2)(x-1)dy = 0 Originally Posted by Archie So it looks nicer. Or perhaps he integrated $\displaystyle (x-1)$ as a single term. You could even do substitutions before you start.
can you explain further?

1dx, x2y, y2x1dy 