# What is a particular solution to the differential equation dy/dx=2xsqrt(1-y^2) and y(0)=0?

Sep 7, 2016

$y = \sin {x}^{2}$

#### Explanation:

this is separable

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

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

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

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

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

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

using the IV, $y \left(0\right) = 0$

$0 = \sin \left(0 + C\right) \implies C = 0$

meaning

$y = \sin {x}^{2}$