Solve differential equation with variation of parameters

And your question is?

It should be easy to see that e^x and e^-x are two independent solutions to y"- y= 0.

The "variation of parameters" is to look for a solution of the form
y(x)= u(x)e^x+ v(x)e^-x
In fact, there are an infinite number of u(x), v(x) that would work.

Differentiating, y'= u'(x)e^x+ u(x)e^x+ v'(x)e^-x- v(x)e^-x.

ASSUME that u'(x)e^x+ v'(x)e^-x= 0.
(Since there are an infinite number of u(x), v(x) that would work above, this is just "narrowing the search".)

With that assumption y'= u(x)e^x- v(x)e^-x.

Differentiating again, y"= u'(x)e^x+ u(x)e^x- v'(x)e^-x+ ve^-x.

Plug that into the original equation and, because e^x and e^-x satisfy the original homogeneous equation, the "u(x)" and "v(x)" terms cancel leaving just u'e^x- v'e^-x= 2/(1+e^x). Treat that, along with
u'e^x+ v'e^-x= 0 (above) as two linear equations for u', v'.
Integrate those solutions to find u and v and plug into y(x)= u(x)e^x+ v'(x)e^-x.