Step 1: Understanding the Question:
We are given a first-order differential equation and an initial value condition.
We need to find the particular solution.
Step 2: Key Formula or Approach:
This equation is of the variable separable type. We rearrange terms to group \(y\) terms with \(dy\) and \(x\) terms with \(dx\).
Standard integral: \(\int \frac{1}{1+t^2} \, dt = \tan^{-1}t + C\).
Step 3: Detailed Explanation:
1. Separate the variables:
\[ \frac{1}{1+y^2} \, dy = \frac{1}{1+x^2} \, dx \]
2. Integrate both sides:
\[ \int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2} \]
\[ \tan^{-1}y = \tan^{-1}x + C \]
3. Find the constant \(C\) using \(y(0)=1\):
Substitute \(x = 0\) and \(y = 1\):
\[ \tan^{-1}(1) = \tan^{-1}(0) + C \]
Since \(\tan^{-1}(1) = \frac{\pi}{4}\) and \(\tan^{-1}(0) = 0\):
\[ \frac{\pi}{4} = 0 + C \Rightarrow C = \frac{\pi}{4} \]
4. Write the final equation:
\[ \tan^{-1}y = \tan^{-1}x + \frac{\pi}{4} \]
Step 4: Final Answer:
The solution to the differential equation is \(\tan^{-1}y = \tan^{-1}x + \frac{\pi}{4}\).