# Find the solution of the differential equation  x(1+y^2)dx-y(1+x^2)dy =0?

Oct 6, 2017

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

#### Explanation:

We have:

$x \left(1 + {y}^{2}\right) \mathrm{dx} - y \left(1 + {x}^{2}\right) \mathrm{dy} = 0$

If we rearrange this ODE from differential form into standard form we have:

$y \left(1 + {x}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = x \left(1 + {y}^{2}\right)$

$\therefore \frac{y}{1 + {y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{1 + {x}^{2}}$

This is now a separable ODE, do we can "separate the variables" to give:

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

We can manipulate the numerators on both sides to get:

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

Both integrals are identical and of standard functions, so we can now integrate to get the General Solution:

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

We are given that $y = 0$ when $x = 1$, so:

$\ln | 1 | = \ln | 1 + 1 | + C + > C = - \ln 2$

So we have the Particular Solutions

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

$\therefore \ln \left(1 + {y}^{2}\right) = \ln \left(\frac{1 + {x}^{2}}{2}\right)$

$\therefore 1 + {y}^{2} = \frac{1 + {x}^{2}}{2}$

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

$\therefore {y}^{2} = \frac{{x}^{2} - 1}{2}$