# What is a solution to the differential equation x+2ysqrt(x^2+1)dy/dx with y(0)=1?

Aug 1, 2016

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

#### Explanation:

IF its actually $x + 2 y \sqrt{{x}^{2} + 1} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

then we can say that

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

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

Integrating both sides

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

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

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

$\implies {y}^{2} = - \sqrt{{x}^{2} + 1} + C$

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

$y = \pm \sqrt{C - \sqrt{{x}^{2} + 1}}$

applying the IV: $y \left(0\right) = 1$

$1 = \pm \sqrt{C - \sqrt{1}}$

so we can disregard the -ve sign

$1 = \sqrt{C - 1} \implies C = 2$

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