Question

Differentials Problem... Possibly solvable with Laplace Transforms?
y''-y=1/(1+e^(t))

Answer 1

I also have y(0)=0 and y'(0)=1+ln(2)

Answer 2

which is what makes me think it's a laplace problem.

Answer 3

it's just that this part here:
1/(1+e^(t))
isn't in my table of transforms.

Answer 4

clearly that's the tricky part here

Answer 5

that and also the y'(0)=1+ln(2), which makes me think those two parts were just made for each other.

Answer 6

well there is always variation of parameters I guess

Answer 7

\[y''-y=0\implies y_c=c_1e^t+c_2e^{-t}\]then you need the wronskinan...
that should work, right?

Answer 8

i have no idea about wronskians. we completely skipped that section in class.

Answer 9

well then I guess I'm not using the method they wanted you to use...

Answer 10

the formula for the wronskian is\[W=\left|\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right|\]here's a link on the technique:
http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

Answer 11

do you use the same y1 and y2 as you would for variation of parameters?

Answer 12

and would those be y1=sint and y2=cost?

Answer 13

no, there are two real roots to the characteristic equation, so the solution is as I wrote above.
y1=e^t
y2=e^(-t)

Answer 14

I love this site! Thank you so much for your help <3
I took the variation of parameters route and got y"-y=v1'e^t-v2'e^(-t)
Solved for v1'e^t=-v2'e^(-t)
Plugged back into original problem
-2v2'e^(-t)=1/(1+e^t)
And ran into that horrible thing again on the right hand side :(

Answer 15

but you didn't use the right formula for variation of parameters\the particular is #38 on this chart
what they call f(t) is what is on the right side, so in your case it's f(t)1/(1+e^(t))
the integrals after multiplying that by y1 and y2 and dividing by the wronskian should hopefully be doable.
just make sure you use the formula correctly and it should work out
You're welcome, and good night :)