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Variable substitution question(diff)

  1. Aug 16, 2009 #1
    [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

    ??
     
  2. jcsd
  3. Aug 19, 2009 #2
    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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