Solve each of the following differential equations
`sqrt((1-x^2)(1-y^2))dy+xydx=0`.

Answer

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

### Step 1: Rearrange the equation
We start by rearranging the given equation:
\[ \sqrt{(1-x^2)(1-y^2)} dy = -xy dx \]

### Step 2: Separate variables
Next, we separate the variables \(x\) and \(y\):
\[ \frac{dy}{\sqrt{1-y^2}} = -\frac{xy}{\sqrt{1-x^2}} dx \]

### Step 3: Introduce substitutions
To simplify the integration, we can use the substitutions:
- Let \(t = \sqrt{1 - y^2}\) which implies \(y = \sqrt{1 - t^2}\) and \(dy = -\frac{t}{\sqrt{1 - t^2}} dt\).
- Let \(u = \sqrt{1 - x^2}\) which implies \(x = \sqrt{1 - u^2}\) and \(dx = -\frac{u}{\sqrt{1 - u^2}} du\).

### Step 4: Substitute into the equation
Substituting into the equation gives:
\[ -\frac{t}{\sqrt{1 - t^2}} dt = -\frac{xy}{\sqrt{1 - x^2}} \left(-\frac{u}{\sqrt{1 - u^2}} du\right) \]
This simplifies to:
\[ \frac{t}{\sqrt{1 - t^2}} dt = \frac{xy}{\sqrt{1 - x^2}} \frac{u}{\sqrt{1 - u^2}} du \]

### Step 5: Integrate both sides
Now, we integrate both sides. The left-hand side becomes:
\[ \int \frac{t}{\sqrt{1 - t^2}} dt = -\sqrt{1 - t^2} + C_1 \]
And the right-hand side becomes:
\[ \int \frac{u}{\sqrt{1 - u^2}} du = -\sqrt{1 - u^2} + C_2 \]

### Step 6: Combine results
After integrating, we can combine the results:
\[ -\sqrt{1 - t^2} = -\sqrt{1 - u^2} + C \]
Where \(C = C_1 - C_2\).

### Step 7: Substitute back for \(t\) and \(u\)
Substituting back \(t = \sqrt{1 - y^2}\) and \(u = \sqrt{1 - x^2}\), we get:
\[ -\sqrt{1 - (1 - y^2)} = -\sqrt{1 - (1 - x^2)} + C \]
This simplifies to:
\[ y = x + C \]

### Final Solution
Thus, the solution to the differential equation is:
\[ \sqrt{1 - y^2} + \frac{1}{2} \ln \left| \frac{\sqrt{1 - y^2} - 1}{\sqrt{1 - y^2} + 1} \right| = \sqrt{1 - x^2} + C \]