Variable substitution question(diff)

[proto]

\frac{xy'}{(\ln x\arctan y)-1}=(1+y^2)\arctan y\\
t=\arctan y\\
t'=\frac{1}{1+y^2}y'\\
\frac{x}{\ln (x)t-1}=\frac{t}{t'}\\

\frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\
xdt=(\ln (x)t-1)tdx\\
\frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\
still can't beak it as one type of variable on each side
so i substitute by another variable
z=\ln x\\
dz=\frac{dx}{x}\\
\frac{dt}{zt-1}=dz\\
i don't know how to separate each variable type on one side
so i could integrate

??

 

[foxjwill]

First of all, your last substitution should result in

t'+t=zt^2​

I'm not exactly sure on a good substitution to use, but since this is a http://mathworld.wolfram.com/BernoulliDifferentialEquation.html" , you can use that method.