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

sciencesolve | Teacher | (Level 3) Educator Emeritus

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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)).`

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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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 }