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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