Question

Am I solving this DE correctly?

Answer 1

I am just tryng some substitution problems.

y' = ln(y+x)

Answer 2

(this particular prob, I made up)

z = y + x
z' = y' +1
y' = z' - 1

z' - 1 = ln(z)
z' = ln(z) + 1
1/ (ln(z) + 1) • (dz/dx) = 1

then I integrate both sides with respect to x, and get:

Integral, 1/ (ln(z) + 1) dz= x

Answer 3

I guess that should work, you got a separable right?

Answer 4

Actually, I don't really know what the Ei here is, wolfram gave this http://www.wolframalpha.com/input/?i=y%27%3Dln%28y%2Bx%29 maybe @zepdrix @IrishBoy123 can help you with this one

Answer 5

After the v sub, it is separable, but then I have an obstacle when it comes to integrating.

Well, this isn't something that I so far really know how to integrate....
Wolfram gives me:

\(\color{black}{\dfrac{{\rm Ei}(\log(z)+1)}{e}}\)

Answer 6

It would be good if someone explains what this Ei really is.

Answer 7

Oh it says "Ei(x) is the exponential integral Ei" ok still no idea haha

Answer 8

Yes, that is my prob as well.

Answer 9

I guess I will just take a note of this for now, and will dig into it later.... because I don't think I have a proper base knowledge for this. It seems too complex. I am better start of some normal examples on the web....

didn't mean to disturb anyone.

Answer 10

https://en.wikipedia.org/wiki/Exponential_integral

Answer 11

@SithsAndGiggles

Answer 12

@freckles

Answer 13

Irishboy, it's ok.

Answer 14

Thank you though

Answer 15

@Zarkon

Answer 16

looks like:
\[\int\limits_{}^{} \frac{e^x}{x} dx =Ei(x)+C \\ \int\limits \frac{1}{\ln(v)+1} dv \\ u=\ln(v)+1 \\ u-1=\ln(v) \\ e^{u-1}=v \\ e^{u-1} du=dv \\ \int\limits \frac{1}{u} e^{u-1} du =\frac{1}{e} \int\limits \frac{e^u}{u} du=\frac{1}{e} Ei(u)+C \\ =\frac{1}{e} Ei(\ln(v)+1)+C\]