Question

Solve the differential equation \( \frac{dy}{dx} = \frac{1+y^2}{1+x^2} \) given \(y(0)=1\).

• A. \( \tan^{-1}y = \tan^{-1}x + \frac{\pi}{6} \)
• B. \( \tan^{-1}y = \tan^{-1}x + \frac{\pi}{4} \)
• C. \( \tan^{-1}y = \tan^{-1}x + \frac{\pi}{3} \)
• D. \( \tan^{-1}y = \tan^{-1}x + \frac{\pi}{2} \)

Hint:

If a differential equation can be written in the form \[ f(y)\,dy=g(x)\,dx \] then it is a separable differential equation and can be solved by integrating both sides.

Answer 1

The Correct Option is B

Concept: The given differential equation is separable. We separate the variables \(x\) and \(y\) and integrate both sides.

Step 1: Separate the variables. \[ \frac{dy}{dx}=\frac{1+y^2}{1+x^2} \] \[ \frac{dy}{1+y^2}=\frac{dx}{1+x^2} \]

Step 2: Integrate both sides. \[ \int \frac{dy}{1+y^2}=\int \frac{dx}{1+x^2} \] \[ \tan^{-1}y=\tan^{-1}x+C \]

Step 3: Apply the initial condition. Given \(y(0)=1\): \[ \tan^{-1}(1)=\tan^{-1}(0)+C \] \[ \frac{\pi}{4}=0+C \] \[ C=\frac{\pi}{4} \] Thus the required solution is \[ \boxed{\tan^{-1}y=\tan^{-1}x+\frac{\pi}{4}} \]