## idku one year ago Am I solving this DE correctly?

1. idku

I am just tryng some substitution problems. y' = ln(y+x)

2. idku

(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

3. Astrophysics

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

4. Astrophysics

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

5. idku

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

6. idku

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

7. Astrophysics

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

8. idku

Yes, that is my prob as well.

9. idku

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.

10. IrishBoy123
11. IrishBoy123

@SithsAndGiggles

12. IrishBoy123

@freckles

13. idku

Irishboy, it's ok.

14. idku

Thank you though

15. Astrophysics

@Zarkon

16. freckles

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