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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

t - 5

f (x) = \frac{8}{x ^{2}}, at x = 2

x \to 0 lim (\frac{2}{x + 5})

x^{2} + y^{2} = c^{2}

x \to - 2 lim (\frac{x ^{2} + 5 x + 6}{x})

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