Question

How did they solve for "y"?

Answer 1

I'm trying to rationalize how they solved for 'y' But I don't see how they got the final answer: \[\frac{ 1 }{ 4 }\ln ( \frac{ y-2 }{ y+2 }) = x + C\]I was trying to see how they got it in the form: \[y = 2 \frac{1+ce^{4x} }{ 1-ce^{4x} }\]

This is as far as I got:
I took the exponent and got rid of the natural log from the left side: \[e^{\ln(\frac{ y-2 }{ y+2}) } = e^{4x+c}\]
to get: \[\frac{ y-2 }{ y+2 } = \pm e^{4x+c}\]
but from here, I'm not sure how they got their answer the form I put earlier.

Answer 2

call e^c a new constant (they called it C, but that is confusing)
for ease of typing C e^4x = B
we have
(y-2)/(y+2) = B
y-2 = By + 2B
y-By = 2+2B
y(1-B)= 2(1+B)
y=2 (1+B)/(1-B)