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

Answer

To solve the initial value equation

\[ \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \]

with the initial condition \( y(0) = 1 \), we will follow these steps:

### Step 1: Separate the Variables
We can rewrite the equation in a separable form:

\[ \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2} \]

### Step 2: Integrate Both Sides
Now, we will integrate both sides:

\[ \int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2} \]

The integral of \(\frac{1}{1 + y^2}\) is \(\tan^{-1}(y)\) and the integral of \(\frac{1}{1 + x^2}\) is \(\tan^{-1}(x)\):

\[ \tan^{-1}(y) = \tan^{-1}(x) + C \]

### Step 3: Apply the Initial Condition
We know that \(y(0) = 1\). We can use this to find the constant \(C\):

\[ \tan^{-1}(1) = \tan^{-1}(0) + C \]

Since \(\tan^{-1}(1) = \frac{\pi}{4}\) and \(\tan^{-1}(0) = 0\), we have:

\[ \frac{\pi}{4} = 0 + C \implies C = \frac{\pi}{4} \]

### Step 4: Write the General Solution
Now we can substitute \(C\) back into the equation:

\[ \tan^{-1}(y) = \tan^{-1}(x) + \frac{\pi}{4} \]

### Step 5: Use the Tangent Addition Formula
Using the tangent addition formula, we can express this as:

\[ \tan^{-1}(y) - \tan^{-1}(x) = \frac{\pi}{4} \]

This implies:

\[ \tan\left(\tan^{-1}(y) - \tan^{-1}(x)\right) = 1 \]

Using the formula for \(\tan(A - B)\):

\[ \frac{y - x}{1 + xy} = 1 \]

### Step 6: Rearranging the Equation
From this, we can rearrange to find \(y\):

\[ y - x = 1 + xy \]

Rearranging gives:

\[ y - xy = 1 + x \]

Factoring out \(y\):

\[ y(1 - x) = 1 + x \]

Thus, we have:

\[ y = \frac{1 + x}{1 - x} \]

### Final Solution
The solution to the initial value problem is:

\[ y = \frac{1 + x}{1 - x} \]