What is the solution for (1+y^2)dx-(x^2+1)dy=0 ?

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

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(1+y^2)dx-(x^2+1)dy=0

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

y = tan (arctan (x) - c_{1})

Solution steps

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

Solve separable ODE:

y = tan (arctan (x) - c_{1})

y = tan (arctan (x) - c_{1})

\frac{\partial}{\partial x} (\frac{- y ^{4}}{4 x ^{4}})

\int (\frac{1}{1 - x}) d x

f (x) = 2 x^{2} + 2 x, (- 2, - 6)

\int_{0}^{2} ln (1 + x^{4}) d x

\int 3^{- x} d x

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