91Ó°ÊÓ

The given differential equation is $x\sqrt{1-y^2}dx+y\sqrt{1-x^2}dy=0$.

The correlation between the variables xand y, which is obtained after removing the derivatives where the relation includes arbitrary constants to represent the order of an equation, is the general solution of the differential equation and a unique solution of the type$\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$that satisfies the differential equation is known as a particular solution. The general solution of the differential equation is used to derive the particular solution by giving values to the arbitrary constants.

Solve the given differential equation$x\sqrt{1-y^2}dx+y\sqrt{1-x^2}dy=0$ using the separable method.

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

Integrate both sides to get the general solution of this differential equation.

$$\begin{gathered} ∫\frac{x}{\sqrt{1-x^2}}dx=-∫\frac{y}{\sqrt{1-y^2}}dy \\ -\sqrt{1-x^2}+C=\sqrt{1-y^2} \\ 1-x^2-2C\sqrt{1-x^2}+C^2=1-y^2 \end{gathered}$$

Therefore, the general solution is$y=\sqrt{x^2+2c\sqrt{1-x^2-C^2}}$ .

Now find the particular solution by applying the boundary condition$y=\frac{1}{2}$ when$x=\frac{1}{2}$ .

$$\begin{gathered} \frac{1}{2}=\sqrt{{\left(\frac{1}{2}\right)}^2+2C\sqrt{1-{\left(\frac{1}{2}\right)}^2}-C_2} \\ \frac{1}{4}=\frac{1}{4}+2C\sqrt{1-\frac{1}{4}}-C_2 \\ C=\sqrt{3} \end{gathered}$$

So, the particular solution is$Y=\sqrt{x^2+\sqrt{12(1-x^2)}-3}$ .

Now, find the slope for sketching the slope field. For this, deriving the general solution, or by rewriting the differential equation itself by putting the first derivative in one side, and everything else in another side.

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

Therefore, the general solution is$y=\sqrt{x^2+2C\sqrt{1-x^2}-C^2}$ and the particular solution is$y=\sqrt{x^2+\sqrt{12(1-x^2)}-3}.$ .