Problem Statement
Find the particular solution of the differential equation:
$$(1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}$$
given that $y = 0$ when $x = 1$.
Verified Solution & Marking Scheme
Divide by (1 + x²) to obtain standard linear form
$$\frac{dy}{dx} + \left(\frac{2x}{1 + x^2}\right)y = \frac{1}{(1 + x^2)^2}$$
This is of the form $$\frac{dy}{dx} + P(x)y = Q(x)$$, where $$P(x) = \frac{2x}{1 + x^2}$$ and $$Q(x) = \frac{1}{(1 + x^2)^2}$$.
Find Integrating Factor (I.F.)
$$\text{I.F.} = e^{\int P(x) \, dx} = e^{\int \frac{2x}{1 + x^2} \, dx} = e^{\ln(1 + x^2)} = 1 + x^2$$
Find general solution
$$y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$$
$$y(1 + x^2) = \int \frac{1}{(1 + x^2)^2} (1 + x^2) \, dx + C = \int \frac{1}{1 + x^2} \, dx + C = \tan^{-1} x + C$$
Apply boundary condition y(1) = 0
Substitute $$x = 1, y = 0$$:
$$0(1 + 1^2) = \tan^{-1}(1) + C \implies 0 = \frac{\pi}{4} + C \implies C = -\frac{\pi}{4}$$
$$y(1 + x^2) = \tan^{-1} x - \frac{\pi}{4} \implies y = \frac{\tan^{-1} x - \frac{\pi}{4}}{1 + x^2}$$