a little help with my differential eqaution?

Question

can someone pls answer this for me using method of variation of parameter? (D^2 - 1) = 2e^(-x) (1+e^(-2x))^(-2) ; where D is the differential operator d/dx

(D^2 - 1) = 2e^(-x) (1+e^(-2x))^(-2)
(m^2 -1) = 0
therefore, m1=1, m2= -1.
therefore, Ae^x + Be^(-x) = yc
therefore, A(x)e^x + B(x)e^(-x) = yp
yp' = A'(x)e^x + A(x)e^x + B'(x)e^(-x) - B(x)e^(-x)
let: A'(x)e^x + B'(x)e^(-x) =0; eq1
yp' = A(x)e^x - B(x)e^(-x)
yp'' = A'(x)e^x + A(x)e^x - B'(x)e^(-x) + B(x)e^(-x)
(D^2 - 1)yp=2e^...
A'(x)e^x - B'(x)e^(-x) = 2e^(-x) (1+e^(-2x))^(-2) ; eq2
solve simultaneously eq1&eq2:
A'(x)e^x = e^(-x) (1+e^(-2x))^(-2)
B'(x)e^(-x) = -e^(-x) (1+e^(-2x))^(-2)
therefore,
A'(x)= e^(-2x) (1+e^(-2x))^(-2)
B'(x)= -(1+e^(-2x))^(-2)
by integrating:
A = 0.5(1+e^(-2x))^(-1)
B = then this i can't integrate.
the answer btw is y = yc -xe^(-x) -0.5e^(-x) ln(1 + e^(-2x))

Answer 1

See answer to your previous version of this question.

Also, see line 8 of
http://www.sosmath.com/tables/integral/integ27/integ27.html

Answer 2

i can help u with it....but alteast post some work of urs......and tell me wheres the difficulty........u cant expect some one do all the work for you......

i am working on it....just gimme some time.....

ur work is correct .....now u have integrate B`(x)
for which try this substitution
put e^-2x =t
u ll get an inetgral of the form (2*t*(1+t)^2) ^-1
or integral of ( 1/(2*t*(1+t)^2) )
which can be integrated by partial fractions method


u do partial fraction integration right? which is done as
a1/t +a2/(1+t) + a3/(1+t)^2

eaach of which can be integrated separately i think.......if u still ahve any problems write to me......

Answer 3

sorry