calculus help!!?

Question

can someone solve this? (D^2 - 1)y = 2e^(-x) (1+e^(-2x))^(-2) ; where D is the differential operator d/dx. thanks

i've tried answering it but i end up getting a different answer from the book and always end up with the same answer whatever method i use. i also tried working backwards from the answer from the book but i dont end up with (D^2 + 1)y=2e^...

the answer btw is y = yc -xe^(-x) -0.5e^(-x) ln(1 + e^(-2x))

Answer 1

The answer you quoted is correct.

proof:
y'' - y = f(x)

f(x) = 2exp(-x) / (1+exp(-2x))^2


find the homogeneous solution yh(x), then the particular solution yp(x) and add them.

y(x) = yh(x) + yp(x)

Homogeneous solution:
y'' - y = 0

yh(x) = A*exp(x) + B*exp(-x)

Particular solution:
yp(x) = u1(x)*exp(x) - u2(x)*exp(-x)

u1(x) = integral(f(x)*(1/2)*exp(-x)dx)
u2(x) = integral(f(x)*(-1/2)*exp(x)dx)

working...
u1(x) = (1/2) / (1+exp(-2x))
u2(x) = (-1/2)*[1 / (1+exp(-2x)) + ln(exp(-2x) / (1+exp(-2x)))]

yp(x) = [(1/2)*(exp(x) + exp(-x)) / (1 + exp(-2x))] + [(1/2)*exp(-x)*ln(exp(-2x) / (1+exp(-2x)))]

yuk! simplifying

yp(x) = exp(x)/2 - [2x + ln(1+exp(-2x))] * exp(-x)/2

y(x) = yh(x) + yp(x)

= A*exp(x) + B*exp(-x) + 0.5*exp(x) - x*exp(-x) - 0.5*exp(-x)*ln(1+exp(-2x))

= C*exp(x) + B*exp(-x) - x*exp(-x) - 0.5*exp(-x)*ln(1+exp(-2x))

y(x) = yh(x) - x*exp(-x) - 0.5*exp(-x)*ln(1+exp(-2x))

Answer 2

I took undergrad differential equations and I'll try my best to solve it. Almost looks like a college differential equation rather than a calculus question. What kind of calculus are you taking??? Sounds like you are in Pre-AP or something for the one's with extra smarts. If this is a calculus question, this is the way to solve it from the best of my knowlege. If it's meant to be a differential equation then it will have to be solved for y. Which I don't think the case is here.

First, it looks like the question asks to separate D first. Because I don't see any y' or y". So this apparently isn't a first or second order differential equation. Also don't see any D' or D''. So looks like a basic derivative problem. Unless you meant D^2 as D". then we have a completly different answer. don't see that as the case though.

Let's solve for D. Wont' show all steps. Too much typing! But did it on anotheer sheet of paper.

This is expression is stated as the square root. I'll use to the ^1/2 power since my comp doesn't have a square root sign.

D = {(2e^(-x) [(1 + e^(-2x)]^-2}^1/2

Now, find the derivative respected to x: After I removed the sq roots and simplfied it down this.

2e^(-x)^1/2 / [ 1 + e^(-2x)] = D

Now you can easily use the quotient rule to solve this and you will have D. Just find the derivative respected to x.
I did this all out of paper again so I won't show all steps. You might want to double check the derivative calculation in case I made a blunder.
D = d/dx = [1 + e^(-2x)]* 1/2[2e^(-x)]^-1/2 - 2e^(-x)^1/2*[-2e^-2x] all divided by: [1 + e^(-2x)]^2.

this was done by the
quotient rule. Hope this helps. And good luck with the calculus class! Calculus rocks!!!!!!!!

Answer 3

i'm not used to do this complex differentiation since i'm in the 11th standard
there's an indian website called www.ilovemaths.com or www.ilovmaths.com
u go to that and there will be an answer waiting for u dear!

Answer 4

The idea is that you do it yourself, what are you going to learn if someone solve the problem for you.

Answer 5

If you do it yourself you'll never need to ask questions like this. I did. My kids hate me.