Question

In Problems 13 to 15, find a solution (or solutions) of the differential equation not obtainable by specializing the constant in your solution of the original problem. Hint: See Example 3.

13. Problem 2

Short answer

Answer

$y=1,y=-1,x=1$, and $x=-1$

Step-by-step solution

Given information

We have given a differential equation $x\sqrt{1-y^2}dx+y\sqrt{1-x^2}dy=0$with the boundary condition $y=\frac{1}{2}$when$x=\frac{1}{2}$.

Definition of a separable differential equation

Any equation of the form $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{f}\left(x\right)\mathit{g}\left(y\right)$is called separable; that is any equation in which dx and terms involving x can be put on one side and dy and terms involving y on the other. For example, $\mathit{f}\left(x\right)\mathit{d}\mathit{x}\mathbf{=}\mathit{g}\mathbf{}\left(y\right)\mathit{d}\mathit{y}$.

Separate the given differential equation

When we separate the variables in the differential equation, we will get

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

Divide the left-hand side by $\sqrt{1-y^2}$and the right-hand side by $\sqrt{1-x^2}$

Now, this step is not valid if $y=1,y=-1,x=1$, and $x=-1$, which are solutions for this differential equation that are not obtained by any choice of C.