# How do you find the particular solution to ysqrt(1-x^2)y'-xsqrt(1-y^2)=0 that satisfies y(0)=1?

Mar 7, 2017

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

#### Explanation:

Grouping variables

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

so it is a separable differential equation.

We have also

$\int \frac{x \mathrm{dx}}{\sqrt{1 - {x}^{2}}} = - \sqrt{1 - {x}^{2}}$ and

$\int \frac{y \mathrm{dy}}{\sqrt{1 - {y}^{2}}} = - \sqrt{1 - {y}^{2}}$ so

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

and $y \left(0\right) = 1$ so

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

so

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