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?


[tex] { y' = (x-3)(y^(2)-2 ), y(3) = 6 } [/tex]


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


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


hence:


[tex] (1/2)ln|(1+y)/(1-y)| - (1/2)ln(2) = x^(2)/2 -3x [/tex]


some counts and we obtain:


[tex] y = (2(e^(x^(2)-6x)))/(1-2(e^(x^(2)-6x))) [/tex]


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


[tex] (2(e^(x^(2)-6x)))/(1-2(e^(x^(2)-6x))) != -1 [/tex]


it should be true that:


[tex] e^(x^(2)-6x) = -1/4 [/tex]


therefore this condition does not create problems.


It follows [tex] y != 1 [/tex] and it should not be true that:


[tex] e^(x^(2)-6x) = 1/4 [/tex]


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


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


[tex] e^(x^(2)-6x) != 1/2 [/tex]


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


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:


[tex] x belongs to ( 3 - sqrt(9 - ln2) , 3 + sqrt(9 -ln4) ) [/tex]


Is everything correct?
Thanks!