Question

Find a solution to the given initial-value problem using Green's functions

Answer 1

Find a solution to the given initial-value problem using Green's functions

$$y"+3y'+2y= \frac{1}{1+e^x}; y(0)=0,y'(0)=1$$

So I have figured out that

$$y_c=e^{-x}-e^{-2x}$$

Now I am having issues figuring out what $y_p$ is. This is what I have so far:

$$y_1=e^{-x}, y_2=e^{-2x},$$

so

$$W(y_1, y_2)=-e^{-3x}.$$

$$G(x,t)=\frac{e^{-t}e^{-2x}-e^{-x}e^{-2t}}{-e^{-3x}} = -e^{-t}e^{x}+e^{-2t}e^{2x}$$

$$y_p=\int^x_0 G(x,t)f(t)dt$$


$$y_p=-e^x \int^x_0 \frac{e^{-t}}{1+e^t}dt+e^{2x} \int^x_0 \frac{e^{-2t}}{1+e^t}dt$$

So

$$y_p=-\ln(e^{-x}+1)(e^x+e^{2x})-\frac{e^{2x}}{2}+\frac{1}{2}+(e^{x}+e^{2x})\ln(2)$$

Is this correct?

Answer 2

I am getting stuck integrating. Like the last 2 lines

Answer 3

I used wolfram to integrate. Idk this function totally scares me

Answer 4

Well, I haven't done much with Green's Function, but when I just plug the whole equation into wolfram, it doesn't seem to agree.

Answer 5

I do remember that you should have something that looks like \[c_1e^{-x}+c_2e^{-2x}+e^{-x}f(x)+e^{-2x}g(x)\]though.

Answer 6

Ya that is more or less the answer but how come I cant come to that answer when integrating :(

Answer 7

aww u got y(0) AND Y'(0) if only they let u use laplace

Answer 8

Have not learned Laplace yet

Answer 9

Wolfram isn't working for me, and I don't have immediate access to mathematica :(

Answer 10

Guess what?!?!? I will fudge the answer :)

Answer 11

hmm i dont rememer that formula for y particular.. are u sure u are doing that part right they usually set it up so that part simplifies

Answer 12

Well, it's cloe to what wolfram is getting, but some signs seem to be disappearing/changing.

Answer 13

did u double check the formula and wronskin just in case

Answer 14

The Wronskin is really simple so that I def got right

Answer 15

okay then i guess ur using parts

Answer 16

Everything seems right to me. It just has a few things go weird at the end.

Answer 17

Yaaaa I was using scraps of paper so I may have messed up. Gonna go try again

Answer 18

Maybe I will just keep it in the integral form and not bother integrating

Answer 19

Even if a few signs are mixed up I think I will still land up with something similar to this. Something ugly looking

Answer 20

Yeah, it'll definitely still be ugly.

Answer 21

Ok so then I guess I will leave it in the integral form

Answer 22

Thanks KG :)

Answer 23

I always can count on u

Answer 24

You're welcome.