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

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

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

\frac{d x}{1 + x ^{2}} + \frac{d y}{1 + y ^{2}} = 0

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

Solution steps

\frac{d x}{1 + x ^{2}} + \frac{d y}{1 + y ^{2}} = 0

Solve separable ODE:

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

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

x cos (y) \frac{d y}{d x} - sin (y) = 0

x^{(- 2)} y^{(- 5)} d x + x^{(- 3)} y^{(- 4)} d y = 0

y^{^{'}} + x - x^{2} \cdot y = x - x^{2}

y^{^{'}} x + 4 y = 1

\frac{d y}{d x} = 6 x^{3} y, y > 0

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