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 DE in standard linear form dy/dx + Py = Q
Dividing through by $(1 + x^2)$:
$$\frac{dy}{dx} + \left( \frac{2x}{1 + x^2} \right) y = \frac{1}{(1 + x^2)^2}$$
Here $P(x) = \frac{2x}{1 + x^2}$ and $Q(x) = \frac{1}{(1 + x^2)^2}$.
Calculate the Integrating Factor (I.F.)
$$\text{I.F.} = e^{\int P \, dx} = e^{\int \frac{2x}{1 + x^2} \, dx} = e^{\ln(1 + x^2)} = 1 + x^2$$
General solution equation y · (I.F.) = ∫ Q · (I.F.) dx + C
$$y(1 + x^2) = \int \frac{1}{(1 + x^2)^2} \cdot (1 + x^2) \, dx + C = \int \frac{1}{1 + x^2} \, dx + C$$
$$y(1 + x^2) = \tan^{-1} x + C$$
Substitute initial condition y = 0 at x = 1
$$0(1 + 1^2) = \tan^{-1}(1) + C \implies 0 = \frac{\pi}{4} + C \implies C = -\frac{\pi}{4}$$
Thus, the particular solution is:
$$y(1 + x^2) = \tan^{-1} x - \frac{\pi}{4} \implies y = \frac{\tan^{-1} x - \frac{\pi}{4}}{1 + x^2}$$