$$-e^{y/x} = ln (1/x) + c $$

can this be solved for y?

Question

$$-e^{y/x} = ln (1/x) + c $$

can this be solved for y?

anonymous · Answer

lets try who knows 
$$e^{\frac{y}{x}}=\ln(x)-c$$
$$\frac{y}{x}=\ln(\ln(x)-c)$$
$$y=\frac{\ln(\ln(x)-c)}{x}$$

unklerhaukus · Answer

yeah i had that but it looked kinda not right

anonymous · Answer

well i made a dumb mistake

anonymous · Answer

$$\frac{y}{x}=\ln(\ln(x)-c)$$
$$y=x\ln(\ln(x)-c)$$

mr.math · Answer

satellite73 just forgot the minus sign there, which will change the expression a bit.

anonymous · Answer

hmm no  i don't think i did

anonymous · Answer

i did make one stupid mistake, let me see if i made another.

mr.math · Answer

Oh I see what you did. Sorry!

anonymous · Answer

no problem, one dumb mistake is my limit!

unklerhaukus · Answer

is this likely to be a answer to a first order homogenous differential equation,
i dont know if i like have the log of a log in my answer

mr.math · Answer

I knew it was a solution of a DE, what was the problem originally?

unklerhaukus · Answer

unklerhaukus · Answer

page above has working to check
the question is
$$xy'=y-xe^{y/x}$$

mr.math · Answer

The solution seems right to me, but I think you missed a minus sign when integrating -1/x.

unklerhaukus · Answer

so the solution has crazy logs of logs , hmm

mr.math · Answer

Yeah, crazy logs are fine too! ;D

unklerhaukus · Answer

well that is good to know
thank you for your help

mr.math · Answer

No problem.