Problem Statement
Solve the differential equation $\frac{dy}{dx} = \frac{y^2 + 1}{x^2 + 1}$, given that $y = 1$ when $x = 0$. Give your answer in the form $y = f(x)$.
Verified Solution & Marking Scheme
Separate Variables
$$\frac{1}{y^2 + 1} \, dy = \frac{1}{x^2 + 1} \, dx$$
Integrate Both Sides
$$\int \frac{1}{y^2 + 1} \, dy = \int \frac{1}{x^2 + 1} \, dx$$
$$\arctan y = \arctan x + C$$
Apply Initial Condition y(0) = 1
$$\arctan(1) = \arctan(0) + C \implies \frac{\pi}{4} = 0 + C \implies C = \frac{\pi}{4}$$
$$\arctan y = \arctan x + \frac{\pi}{4}$$
Take Tangent of Both Sides
$$y = \tan\left(\arctan x + \frac{\pi}{4}\right) = \frac{\tan(\arctan x) + \tan(\pi/4)}{1 - \tan(\arctan x)\tan(\pi/4)} = \frac{x + 1}{1 - x}$$