
#1
Aug1609, 04:12 PM

P: 5

[tex]\frac{xy'}{(\ln x\arctan y)1}=(1+y^2)\arctan y\\[/tex]
[tex]t=\arctan y\\[/tex] [tex]t'=\frac{1}{1+y^2}y'\\[/tex] [tex]\frac{x}{\ln (x)t1}=\frac{t}{t'}\\[/tex] [tex]\frac{x}{\ln (x)t1}=\frac{tdx}{dt}\\[/tex] [tex]xdt=(\ln (x)t1)tdx\\[/tex] [tex]\frac{dt}{\ln (x)t1}=\frac{dx}{x}\\[/tex] still cant beak it as one type of variable on each side so i substitute by another variable [tex]z=\ln x\\[/tex] [tex]dz=\frac{dx}{x}\\[/tex] [tex]\frac{dt}{zt1}=dz\\[/tex] i dont know how to separate each variable type on one side so i could integrate ?? 



#2
Aug1909, 06:50 PM

P: 355

First of all, your last substitution should result in
[tex]t'+t=zt^2[/tex]I'm not exactly sure on a good substitution to use, but since this is a Bernoulli ODE, you can use that method. 


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