Variation of Parameters

**ihmth** · August 6th 2009, 07:57 AM

Using variation of parameters, solve for the general solution of:
$(D^2-2D+1)y = (e^2)^x((e^x+1)^-)^2$

I tried to solve for this several times but I keep getting the wrong answer, so I need help pls.

-tnx

**Calculus26** · August 7th 2009, 08:48 AM

See attachment

Note : in the attachment I wrote the soultion as

c1e^(x) +c2xe^(x) +e^(x)[ln(e^x+1)-x]

however the -x e^(x) can be absorbed into c2xe^(x) so it should read:

y = c1e^x +c2xe^x + e^(x)ln(e^x+1)

**ihmth** · August 8th 2009, 07:33 AM

In my answer I keep getting an extra term " $-xe^x$ " and according in the book the answer is y = yc + $e^xln(1 + e^x)$ ...where yc is the complimentary solution. So it is possible for $-xe^x$ to be absorbed by one of the terms in the complementary solution...Now I know

thank you

**Calculus26** · August 8th 2009, 08:59 AM

I edited my last post to make that exact point as at first I did the same thing

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