# Variable substitution question(diff)

 P: 5 $$\frac{xy'}{(\ln x\arctan y)-1}=(1+y^2)\arctan y\\$$ $$t=\arctan y\\$$ $$t'=\frac{1}{1+y^2}y'\\$$ $$\frac{x}{\ln (x)t-1}=\frac{t}{t'}\\$$ $$\frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\$$ $$xdt=(\ln (x)t-1)tdx\\$$ $$\frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\$$ 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 ??
 P: 355 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.