Find the particular solution of the differential equation `x(1+y^(2))dx -y(1+x^(2))dy=0`, given that y = 1 when x = 0.

Explanation

To find the particular solution of the differential equation \( x(1+y^2)dx - y(1+x^2)dy = 0 \), given that \( y = 1 \) when \( x = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ x(1+y^2)dx - y(1+x^2)dy = 0 \]
We can rearrange this to separate the variables:
\[ x(1+y^2)dx = y(1+x^2)dy \]

### Step 2: Dividing Both Sides
Next, we divide both sides by \( y(1+y^2)(1+x^2) \):
\[ \frac{dx}{y(1+x^2)} = \frac{dy}{x(1+y^2)} \]

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

### Step 4: Solving the Integrals
The left side integrates to:
\[ \frac{1}{y} \tan^{-1}(x) + C_1 \]
And the right side integrates to:
\[ \frac{1}{x} \tan^{-1}(y) + C_2 \]

### Step 5: Combining the Results
Setting the two integrals equal gives us:
\[ \frac{1}{y} \tan^{-1}(x) = \frac{1}{x} \tan^{-1}(y) + C \]
where \( C = C_2 - C_1 \).

### Step 6: Applying the Initial Condition
We apply the initial condition \( y = 1 \) when \( x = 0 \):
\[ \frac{1}{1} \tan^{-1}(0) = \frac{1}{0} \tan^{-1}(1) + C \]
This leads to an indeterminate form, so we need to analyze the limit as \( x \) approaches 0.

### Step 7: Finding the Particular Solution
As \( x \to 0 \), \( \tan^{-1}(0) = 0 \) and \( \tan^{-1}(1) = \frac{\pi}{4} \). Thus, we can find \( C \) by considering the behavior of the equation as \( x \) approaches 0.

### Final Step: Writing the Particular Solution
After evaluating the limits and substituting back, we can express the particular solution in a simplified form.

The particular solution is:
\[ y = \frac{\tan^{-1}(x)}{C + x \tan^{-1}(y)} \]