Solve the following differential equations.
`(dy)/(dx )=(1+y^2)/(1+x^2)` if `y=1` for `x=0`

Answer

To solve the differential equation \(\frac{dy}{dx} = \frac{1+y^2}{1+x^2}\) with the initial condition \(y(0) = 1\), we can follow these steps:

### Step 1: Separate the Variables
We start by separating the variables \(y\) and \(x\):

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

### Step 2: Integrate Both Sides
Next, we 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: Use the Identity for Inverse Tangent
We can use the identity for the difference of inverse tangents:

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

This can be rewritten using the formula:

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

### Step 4: Solve for the Constant \(C\)
To find the constant \(C\), we use the initial condition \(y(0) = 1\):

Substituting \(x = 0\) and \(y = 1\):

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

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

\[ C = \frac{\pi}{4} \]

### Step 5: Substitute Back into the Equation
Now we substitute \(C\) back into our equation:

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

### Step 6: Rearranging the Equation
Rearranging gives us:

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

### Step 7: Use the Tangent Addition Formula
Using the tangent addition formula:

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

Using the identity \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\):

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

### Final Solution
Thus, the particular solution to the differential equation is:

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