solve using method of variation of parameters
y''-y = 2/(1+e^x)
y'' ==> second order
And your question is?
It should be easy to see that ex 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)ex+ v(x)e-x
In fact, there are an infinite number of u(x), v(x) that would work.
Differentiating, y'= u'(x)ex+ u(x)ex+ v'(x)e-x- v(x)e-x.
ASSUME that u'(x)ex+ 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)ex- v(x)e-x.
Differentiating again, y"= u'(x)ex+ u(x)ex- v'(x)e-x+ ve-x.
Plug that into the original equation and, because ex and e-x satisfy the original homogeneous equation, the "u(x)" and "v(x)" terms cancel leaving just u'ex- v'e-x= 2/(1+ex). Treat that, along with
u'ex+ 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)ex+ v'(x)e-x.
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