What is the solution for y^'tan^{-1}(x)-y(1+x^2)^{-1}=0 ?

The solution for y^'tan^{-1}(x)-y(1+x^2)^{-1}=0 is y=c_{1}arctan(x)

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y^'tan^{-1}(x)-y(1+x^2)^{-1}=0

y^{^{'}} tan^{- 1} (x) - y (1 + x^{2})^{- 1} = 0

y = c_{1} arctan (x)

Solution steps

y^{'} (x) arctan (x) - y (1 + x^{2})^{- 1} = 0

Solve separable ODE:

y = c_{1} arctan (x)

y = c_{1} arctan (x)

What is the solution for y^'tan^{-1}(x)-y(1+x^2)^{-1}=0 ?
The solution for y^'tan^{-1}(x)-y(1+x^2)^{-1}=0 is y=c_{1}arctan(x)