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})

\int (ln (x) - ln (x y)) d x

x \to \infty lim (i x)

\frac{d}{d x} (4 x^{3} - 5 x^{2} + 4 x - 1)

x \to - \infty lim (\frac{19}{20 + ( \frac{1}{x} )})

\int_{2}^{4} x e^{\frac{x}{2}} d x

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})