# How can I solve this differential equation? :  xy \ dx-(x^2+1) \ dy = 0

Feb 14, 2018

$y = A \sqrt{{x}^{2} + 1}$

#### Explanation:

we have in differential form:

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

If we put in standard form, and collect terms:

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

Which is a First Order Separable Ordinary Differential Equation, so we can separate the variables to get:

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

We can manipulate the RHS integral as follows:

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

And now both integrals are standard results, so integrating give us:

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

Noting that we require areal solution, and writing $C = \ln A$, we get:

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

Giving us the General Solution:

$y = A \sqrt{{x}^{2} + 1}$