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
Write in standard linear form
Divide throughout by $(1 + x^2)$:
$$\frac{dy}{dx} + \left( \frac{2x}{1 + x^2} \right) y = \frac{1}{(1 + x^2)^2}$$
This is of the form $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{1}{(1+x^2)^2}$.
Compute the 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 the general solution
$$y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$$
$$y(1 + x^2) = \int \frac{1}{(1 + x^2)^2} (1 + x^2) \, dx + C = \int \frac{1}{1 + x^2} \, dx + C$$
$$y(1 + x^2) = \tan^{-1} x + C$$
Apply initial condition y(1) = 0
Substitute $x = 1, y = 0$:
$$0(1 + 1^2) = \tan^{-1}(1) + C \implies 0 = \frac{\pi}{4} + C \implies C = -\frac{\pi}{4}$$
Particular solution:
$$y = \frac{\tan^{-1} x - \frac{\pi}{4}}{1 + x^2}$$