## variable substitution question(diff)

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

$$\frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\$$
$$xdt=(\ln (x)t-1)tdx\\$$
$$\frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\$$
still cant 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 dont know how to separate each variable type on one side
so i could integrate

??
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 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 Bernoulli ODE, you can use that method.