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Lets start with the integration of the partial fractions where we have
\(\displaystyle \frac{1}{2}ln(\frac{|y-1|}{|y+1|}) = ln(|x|)\, +\, C\)
In order to find out what the constant C is, we use the initial condition y=2 when x=1. Substituting those two values into the above equation we get
\(\displaystyle \frac{1}{2}ln(\frac{|2-1|}{|2+1|}) = ln(|1|)\, +\, C\)
or
\(\displaystyle \frac{1}{2}ln(\frac{1}{3}) = C\)
and we can write the solution as
\(\displaystyle \frac{1}{2}ln(\frac{|y-1|}{|y+1|}) = ln(|x|)\, +\, \frac{1}{2}ln(\frac{1}{3})\)
or
\(\displaystyle ln(\frac{|y-1|}{|y+1|}) = 2\, ln(|x|)\, +\, ln(\frac{1}{3})\)
Using a couple of the rules for logarithms this gives
\(\displaystyle ln(\frac{|y-1|}{|y+1|}) = ln(\frac{1}{3}x^2)\)
Lets look at the solution:
First, suppose either y>1 OR y<-1. Then we can remove the absolution value signs since both y-1 and y+1 are positive and the ratio is positive OR both y-1 and y+1 are negative and the ratio is positive. So, again using one of the rules for logarithms, we have
\(\displaystyle \frac{y-1}{y+1} = \frac{1}{3}x^2\)
or, solving for y
\(\displaystyle y = \frac{1\, +\, \frac{1}{3}x^2}{1\, -\, \frac{1}{3}x^2}\, =\, \frac{3\, +\, x^2}{3\, -\, x^2}\)
Next, suppose -1<y<1. Then y-1 and y+1 are of opposite signs and we will need to reverse the sign of one of the expression when removing the absolute value sign. Thus, again using one of the rules for logarithms, we have
\(\displaystyle \frac{1-y}{y+1} = \frac{1}{3}x^2\)
or, solving for y
\(\displaystyle y = \frac{1\, -\, \frac{1}{3}x^2}{1\, +\, \frac{1}{3}x^2}\, =\, \frac{3\, -\, x^2}{3\, +\, x^2}\)