# How do you find the general solution to (x^2+1)y'=xy?

Jul 24, 2016

$y = \sqrt{C \left({x}^{2} + 1\right)}$

#### Explanation:

$\left({x}^{2} + 1\right) y ' = x y$

separate it

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

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

$\int \frac{d}{\mathrm{dx}} \left(\ln y\right) \setminus \mathrm{dx} = \int \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{2} \ln \left({x}^{2} + 1\right)\right) \setminus \mathrm{dx}$

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

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

$y = \sqrt{C \left({x}^{2} + 1\right)}$