Question

(ydy/xdx)+[2(x^2+y^2-1)/(x^2+y^2+1)]=0​

Answer 1

Answer:

$\frac{dy}{dx} = - \frac{2( {x}^{3 } + {y}^{2}x - x)}{ {x}^{2}y + {y}^{3} + y}$

Step-by-step explanation:

$\frac{ydy}{xdx} + \frac{2( {x}^{2 } + {y}^{2} - 1)}{ {x}^{2} + {y}^{2} + 1} = 0 \\ \frac{ydy}{xdx} = - \frac{2( {x}^{2 } + {y}^{2} - 1)}{ {x}^{2} + {y}^{2} + 1} \\ \frac{dy}{dx} = - \frac{2( {x}^{3 } + {y}^{2}x - x)}{ {x}^{2}y + {y}^{3} + y}$