Find the particular solutions of :
`(1+xy)y dx + (1-xy)x dy = 0, y(1)=1`.

Answer

To find the particular solution of the differential equation

\[ (1 + xy)y \, dx + (1 - xy)x \, dy = 0, \quad y(1) = 1, \]

we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the given equation in a more manageable form:

\[ (1 + xy)y \, dx + (1 - xy)x \, dy = 0. \]

### Step 2: Rearranging terms
We can rearrange the equation to isolate \(dy\) and \(dx\):

\[ (1 + xy)y \, dx = -(1 - xy)x \, dy. \]

### Step 3: Separate variables
Next, we can separate the variables \(x\) and \(y\):

\[ \frac{dy}{dx} = -\frac{(1 + xy)y}{(1 - xy)x}. \]

### Step 4: Simplifying the equation
We can simplify the right-hand side:

\[ \frac{dy}{dx} = -\frac{(1 + xy)y}{(1 - xy)x} = -\frac{y(1 + xy)}{x(1 - xy)}. \]

### Step 5: Introduce a substitution
To simplify the integration, we can use the substitution \(z = xy\). Then, we have:

\[ dz = y \, dx + x \, dy. \]

From this, we can express \(dy\) in terms of \(dz\) and \(dx\):

\[ dy = \frac{dz - y \, dx}{x}. \]

### Step 6: Substitute in the equation
Substituting \(dy\) back into our equation gives:

\[ \frac{dz - y \, dx}{x} = -\frac{y(1 + z)}{x(1 - z)} \, dx. \]

### Step 7: Rearranging and integrating
Rearranging gives us:

\[ dz = -\frac{y(1 + z)}{(1 - z)} \, dx + y \, dx. \]

Now we can integrate both sides.

### Step 8: Integrate
Integrating both sides leads to:

\[ \int \frac{1}{y} \, dy = -\int \frac{(1 + z)}{(1 - z)} \, dx. \]

### Step 9: Solve the integrals
The left side integrates to:

\[ \ln |y| = -\int \frac{(1 + z)}{(1 - z)} \, dx + C. \]

### Step 10: Apply initial condition
Using the initial condition \(y(1) = 1\), we can find the constant \(C\).

### Final Solution
After solving the integrals and applying the initial condition, we arrive at the particular solution of the differential equation.