What is a general solution to the differential equation 1+y^2-y'sqrt(1-x^2)=0?

Jul 20, 2016

$y = \tan \left({\sin}^{- 1} x + C\right)$

Explanation:

$1 + {y}^{2} - y ' \sqrt{1 - {x}^{2}} = 0$

first we separate it
$1 + {y}^{2} = y ' \sqrt{1 - {x}^{2}}$

$\frac{1}{1 + {y}^{2}} y ' = \frac{1}{\sqrt{1 - {x}^{2}}}$

then integrate

$\int \frac{1}{1 + {y}^{2}} y ' \setminus \mathrm{dx} = \int \frac{1}{\sqrt{1 - {x}^{2}}} \setminus \mathrm{dx}$

$\int \frac{1}{1 + {y}^{2}} \setminus \mathrm{dy} = \int \frac{1}{\sqrt{1 - {x}^{2}}} \setminus \mathrm{dx}$

these are standard integrals, done using tan and sin subs, which means....

${\tan}^{- 1} y = {\sin}^{- 1} x + C$

$y = \tan \left({\sin}^{- 1} x + C\right)$