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Variable substitution question(diff)

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proto
#1
Aug16-09, 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)t-1}=\frac{t}{t'}\\[/tex]

[tex]\frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\[/tex]
[tex]xdt=(\ln (x)t-1)tdx\\[/tex]
[tex]\frac{dt}{\ln (x)t-1}=\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}{zt-1}=dz\\[/tex]
i dont know how to separate each variable type on one side
so i could integrate

??
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foxjwill
#2
Aug19-09, 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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