## verifying definition interval of given differential equation

Hi everyone,
I need your help to check the method I followed to solve this given differential equation:

I cant actually write powers in LaTex, how do you do that?

${ y' = (x-3)(y^(2)-2 ), y(3) = 6 }$

now, supposing $y != +/- 1$ is it possible to rewrite the first eq. as:

$int[3,y] dy/(y^(2) -1) = int[6,x] (x-3) dx$

hence:

$(1/2)ln|(1+y)/(1-y)| - (1/2)ln(2) = x^(2)/2 -3x$

some counts and we obtain:

$y = (2(e^(x^(2)-6x)))/(1-2(e^(x^(2)-6x)))$

At this point we look at the conditions imposed during the process, namely $y != +/- 1$:
per $y != -1$, we found that, being:

$(2(e^(x^(2)-6x)))/(1-2(e^(x^(2)-6x))) != -1$

it should be true that:

$e^(x^(2)-6x) = -1/4$

therefore this condition does not create problems.

It follows $y != 1$ and it should not be true that:

$e^(x^(2)-6x) = 1/4$

which equation is solved for $x = 3 +/- sqrt(9 -ln4)$.

Highlighted these details, we have a look at the denominator, who requires the condition:

$e^(x^(2)-6x) != 1/2$

in order to determine the local interval of definiton.
This equation leads to exclude $x$ values equal to $x = 3 +/- sqrt(9 - ln2)$.

All these considerations take to write this kind of graph :

hence we deduct that the (max?) largest definition interval in which the solution is defined is:

$x belongs to ( 3 - sqrt(9 - ln2) , 3 + sqrt(9 -ln4) )$

Is everything correct?
Thanks!