Solve the following initial value equations :
`x(1+y^(2))dx-y(1+x^(2))dy=0`, given that `y=1` when `x=0`.

Answer

To solve the initial value equation \( x(1+y^2)dx - y(1+x^2)dy = 0 \) with the initial condition \( y = 1 \) when \( x = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given differential equation:
\[ x(1+y^2)dx - y(1+x^2)dy = 0 \]
We can rearrange this to separate the variables:
\[ x(1+y^2)dx = y(1+x^2)dy \]
Dividing both sides by \( y(1+y^2)(1+x^2) \) gives:
\[ \frac{x}{y(1+x^2)}dx = \frac{1}{1+y^2}dy \]

### Step 2: Substitution
Let:
\[ u = 1 + x^2 \quad \text{and} \quad t = 1 + y^2 \]
Then, we differentiate:
\[ du = 2x \, dx \quad \Rightarrow \quad x \, dx = \frac{du}{2} \]
\[ dt = 2y \, dy \quad \Rightarrow \quad y \, dy = \frac{dt}{2} \]
Substituting these into our equation gives:
\[ \frac{1}{2} \cdot \frac{du}{u} = \frac{1}{2} \cdot \frac{dt}{t} \]

### Step 3: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{du}{u} = \int \frac{dt}{t} \]
This results in:
\[ \log u = \log t + C \]
Substituting back for \( u \) and \( t \):
\[ \log(1 + x^2) = \log(1 + y^2) + C \]

### Step 4: Exponentiating
Exponentiating both sides gives:
\[ \frac{1 + x^2}{1 + y^2} = e^C \]
Let \( e^C = k \), where \( k \) is a constant:
\[ 1 + x^2 = k(1 + y^2) \]

### Step 5: Applying the Initial Condition
Using the initial condition \( y = 1 \) when \( x = 0 \):
\[ 1 + 0^2 = k(1 + 1^2) \quad \Rightarrow \quad 1 = k(2) \quad \Rightarrow \quad k = \frac{1}{2} \]
Substituting \( k \) back into the equation gives:
\[ 1 + x^2 = \frac{1}{2}(1 + y^2) \]
Multiplying through by 2:
\[ 2 + 2x^2 = 1 + y^2 \quad \Rightarrow \quad 2x^2 - y^2 + 1 = 0 \]

### Final Solution
Thus, the particular solution of the given differential equation is:
\[ 2x^2 - y^2 + 1 = 0 \]