Question

Find the general solution of the differential equation \(\dfrac{dy}{dx}=\dfrac{1+y^2}{1+x^2}\).

Hint:

Separate variables as \(\dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}\) and integrate both sides.

Answer 1

Step 1: Verifying by differentiating the proposed solution:
Differentiate \(\tan^{-1}y=\tan^{-1}x+C\) implicitly w.r.t. \(x\): \(\dfrac{1}{1+y^2}\dfrac{dy}{dx}=\dfrac{1}{1+x^2}\).

Step 2: Solving for dy/dx:
\(\dfrac{dy}{dx}=\dfrac{1+y^2}{1+x^2}\), which is exactly the original equation.

Final Answer:
The implicit solution checks out: \(\boxed{\tan^{-1}y=\tan^{-1}x+C}\).