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ODE: Using the fact that Po(x) = 1 is a solution of (1-x^2)y'' - 2xy' = 0, find a second independent solution using the method of reduction of order of a homogeneous equation.
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Let v = y'. Hence v' = y''. So (1-x^2)v' - 2xv = 0 2xv = (1-x^2)(dv/dx) 2x / (1-x^2) dx = (1/v) dv -ln (1-x^2) = ln v + K ln (1-x^2)^-1 = ln (Cv) Cv = 1 / (1-x^2) y' = A / (1-x^2) y' = X / (1-x) + Y / (1+x) Using partial fractions, X = A / 2 and Y = -A / 2 Then integrate y' to get y.
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