Question

solve the differential equation
√1+x^2dx+√1+y^2dy=0
​

Answer 1

Step-by-step explanation:

$\sqrt{1 + {x}^{2} } dx + \sqrt{1 + {y}^{2} } dy = 0$

$\sqrt{1 + {x}^{2} } dx = - \sqrt{1 + {y}^{2} } dy$

$integrating \: both \: sides \\ \int \sqrt{ {x}^{2} + {1}^{2} } dx = - \int \sqrt{ {y}^{2} + {1}^{2} } dy$

We know that

$\int \sqrt{ {x}^{2} + {a}^{2} } dx = \frac{x}{2} \sqrt{ {x}^{2} + {a}^{2} } + \frac{ {a}^{2} }{2} ln |x + \sqrt{ {x}^{2} + {a}^{2} } | + c$

So Applying the above formula

$\frac{x}{2} \sqrt{ {x}^{2} + {1}^{2} } + \frac{ {1}^{2} }{2} ln |x + \sqrt{ {x}^{2} + {1}^{2} } | = \frac{y}{2} \sqrt{ {y}^{2} + {1}^{2} } + \frac{ {1}^{2} }{2} ln |x + \sqrt{ {x}^{2} + {1}^{2} } |$