# how to express y from this> (ln|y|)/2 - (ln|2+y|)/2 = x^2/2 +C

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You should factor out `1/2` such that:

`(1/2)(ln |y|-ln |2+y|) = x^2/2 + c`

Using the logarithmic identities yields:

`(1/2) ln|y/(2+y)| = x^2/2 + c`

`ln|y/(2+y)| = 2*(x^2/2) + c => ln|y/(2+y)| = x^2 + c`

`|y/(2+y)| = e^(x^2+c)`

Since `e^(x^2+c) > 0` , hence `|y/(2+y)| = y/(2+y)`

`y = (2+y)e^(x^2+c)`

You need to open the brackets:

`y = 2e^(x^2+c) + y*e^(x^2+c)`

You need to move the terms that contain y to the left side:

`y - y*e^(x^2+c) = 2e^(x^2+c)`

You need to factor out y such that:

`y(1 - e^(x^2+c)) = e^(x^2+c)`

You need to divide by `(1 - e^(x^2+c))` such that:

`y = (e^(x^2+c))/(1 - e^(x^2+c))`

**Hence, expressing y in terms of x yields `y = (e^(x^2+c))/(1 - e^(x^2+c)).` **

We know

ln(a) - ln(b) = ln(a/b), we will use this

(ln|y|)/2 - (ln|2+y|)/2 = x^2/2 +C

=> ln (|y|/|2+y|) = x^2 + 2 C

=> |y/(2+y)| = Exp[ x^2 + 2 C]

=> |2/y + 1| = Exp[- x^2 - 2 C]

There are two cases:

**2/y + 1 < 0**that

**y < -2**

then

|2/y + 1| = - 2/y - 1

hence

- 2/y - 1 = Exp[- x^2 - 2 C]

=> 2/y = - Exp[- x^2 - 2 C] - 1

**=> y = - 2/ {**Exp[- x^2 - 2 C] + 1 }

Second case:

**2/y + 1 > 0**that

**y > -2**

|2/y + 1| = 2/y + 1

hence

2/y + 1 = Exp[- x^2 - 2 C]

=> 2/y = Exp[- x^2 - 2 C] - 1

**=> y = 2/ {**Exp[- x^2 - 2 C] - 1 }

**So the expression of Y in terms of X will be:**

**y < -2**

**y = - 2/ { Exp[- x^2 - 2 C] + 1 }**

**and for y> -2**

**y = 2/ { Exp[- x^2 - 2 C] - 1 }**