#integration stuff
75 messages · Page 1 of 1 (latest)
ren
and if so, how?
@fathom iris here
no
...
it doesn't have an elementary antiderivative
#helpers-info
how did u do it that fast
to view commands and rules
austin buddy lemme think atleast
tysm
It's a specific type of integral
..
$y'\cdot\ln y=x$
ren
..
It doesn't have an elementary antiderivative
so i got
$y\ln y-y=\frac{x^{2}}{2}+c$
ren
but has it got any antiderivative at all?
not really
guys?
it's seperable
if i name a function, naxtisys epic function and call it the antiderivative of said function, would you accept it?
yeah
havent done DE sorry :(
$e^{\ln y}\left(\ln y-1\right)=\frac{x^{2}}{2}+c$
that's basically what real mathematicians have done for it
ren
oh
you're not going to get a closed form
wait wait wait
im using productlgo
*productlog
so i think i got smth but WA is giving me something else
,w y'ln(y)=x
wow shocking
$e^{\ln y-1}\left(\ln y-1\right)=\frac{x^{2}}{2e}+c$
ren

bro wait
What is your question ren
probably different simplification
$\ln y=1+W\left(\frac{x^{2}}{2e}+c\right)$
ren
$y=e^{1+W\left(\frac{x^{2}}{2e}+c\right)}$
ren