`ysqrt(1-x^(2)) dy + x sqrt(1-y^(2)) dx=0`

Answer

To solve the differential equation \( y \sqrt{1 - x^2} \, dy + x \sqrt{1 - y^2} \, dx = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation:
\[ y \sqrt{1 - x^2} \, dy = -x \sqrt{1 - y^2} \, dx \]

### Step 2: Separating Variables
Next, we separate the variables:
\[ \frac{y \, dy}{\sqrt{1 - y^2}} = -\frac{x \, dx}{\sqrt{1 - x^2}} \]

### Step 3: Integrating Both Sides
Now, we integrate both sides:
\[ \int \frac{y \, dy}{\sqrt{1 - y^2}} = -\int \frac{x \, dx}{\sqrt{1 - x^2}} \]

### Step 4: Using Substitution for Integration
For the left side, we can use the substitution \( u = 1 - y^2 \), which gives \( du = -2y \, dy \) or \( y \, dy = -\frac{1}{2} du \):
\[ \int \frac{y \, dy}{\sqrt{1 - y^2}} = -\frac{1}{2} \int \frac{du}{\sqrt{u}} = -\frac{1}{2} \cdot 2 \sqrt{u} = -\sqrt{1 - y^2} \]

For the right side, we use the substitution \( v = 1 - x^2 \), which gives \( dv = -2x \, dx \) or \( x \, dx = -\frac{1}{2} dv \):
\[ -\int \frac{x \, dx}{\sqrt{1 - x^2}} = -\left(-\frac{1}{2} \int \frac{dv}{\sqrt{v}}\right) = \frac{1}{2} \cdot 2 \sqrt{v} = \sqrt{1 - x^2} \]

### Step 5: Equating the Results
Now we equate the results of the integrations:
\[ -\sqrt{1 - y^2} = \sqrt{1 - x^2} + C \]
where \( C \) is the constant of integration.

### Step 6: Rearranging to Get the Final Solution
Rearranging gives us:
\[ \sqrt{1 - y^2} + \sqrt{1 - x^2} = -C \]
Let \( A = -C \) (a new constant), we can write:
\[ \sqrt{1 - x^2} + \sqrt{1 - y^2} = A \]

### Final Solution
Thus, the solution to the differential equation is:
\[ \sqrt{1 - x^2} + \sqrt{1 - y^2} = A \]