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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

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