# Solve the Differential Equation y’+Sqrt((1-y^2)/(1-x^2))=0

In this tutorial we shall solve a differential equation of the form $y’ + \sqrt {\frac{{1 – {y^2}}}{{1 – {x^2}}}} = 0$ by using the separating the variables method.

The differential equation of the form is given as
$y’ + \sqrt {\frac{{1 – {y^2}}}{{1 – {x^2}}}} = 0$

This differential equation can also be written as
$\begin{gathered} \frac{{dy}}{{dx}} + \sqrt {\frac{{1 – {y^2}}}{{1 – {x^2}}}} = 0 \\ \Rightarrow \frac{{dy}}{{dx}} + \frac{{\sqrt {1 – {y^2}} }}{{\sqrt {1 – {x^2}} }} = 0 \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{{\sqrt {1 – {y^2}} }}{{\sqrt {1 – {x^2}} }} \\ \end{gathered}$

Separating the variables, the given differential equation can be written as
$\frac{1}{{\sqrt {1 – {y^2}} }}dy = – \frac{1}{{\sqrt {1 – {x^2}} }}dx\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Keep in mind that in the separating variable technique the terms $dy$ and $dx$ are placed in the numerator with their respective variables.

Now integrating both sides of the equation (i), we have
$\int {\frac{1}{{\sqrt {1 – {y^2}} }}dy = – \int {\frac{1}{{\sqrt {1 – {x^2}} }}dx} }$

Using the formula of integration $\int {\frac{1}{{\sqrt {1 – {x^2}} }}dx} = {\sin ^{ – 1}}x + c$, we get
$\begin{gathered} {\sin ^{ – 1}}y = – {\sin ^{ – 1}}x + c \\ \Rightarrow y = \sin \left( { – {{\sin }^{ – 1}}x + c} \right) \\ \end{gathered}$

This is the required solution of the given differential equation.