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

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idku
 one year ago
Best ResponseYou've already chosen the best response.0I am just tryng some substitution problems. y' = ln(y+x)

idku
 one year ago
Best ResponseYou've already chosen the best response.0(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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I guess that should work, you got a separable right?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Actually, 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

idku
 one year ago
Best ResponseYou've already chosen the best response.0After 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: \(\color{black}{\dfrac{{\rm Ei}(\log(z)+1)}{e}}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0It would be good if someone explains what this Ei really is.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh it says "Ei(x) is the exponential integral Ei" ok still no idea haha

idku
 one year ago
Best ResponseYou've already chosen the best response.0Yes, that is my prob as well.

idku
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0looks like: \[\int\limits_{}^{} \frac{e^x}{x} dx =Ei(x)+C \\ \int\limits \frac{1}{\ln(v)+1} dv \\ u=\ln(v)+1 \\ u1=\ln(v) \\ e^{u1}=v \\ e^{u1} du=dv \\ \int\limits \frac{1}{u} e^{u1} du =\frac{1}{e} \int\limits \frac{e^u}{u} du=\frac{1}{e} Ei(u)+C \\ =\frac{1}{e} Ei(\ln(v)+1)+C\]
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