Let $$y = y(x)$$ be the solution of the differential equation $$(1 + x^2)\frac{dy}{dx} + y = e^{\tan^{-1}x}$$, $$y(1) = 0$$. Then $$y(0)$$ is

$$(1+x^2)\frac{dy}{dx} + y = e^{\tan^{-1}x}$$. Divide by (1+x²):

$$\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{e^{\tan^{-1}x}}{1+x^2}$$.

IF = $$e^{\int dx/(1+x^2)} = e^{\tan^{-1}x}$$.

$$ye^{\tan^{-1}x} = \int \frac{e^{2\tan^{-1}x}}{1+x^2}dx$$. Let t = tan⁻¹x:

$$= \int e^{2t}dt = \frac{e^{2t}}{2} + C = \frac{e^{2\tan^{-1}x}}{2} + C$$.

y(1) = 0: $$0 \cdot e^{\pi/4} = e^{\pi/2}/2 + C \Rightarrow C = -e^{\pi/2}/2$$.

$$ye^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} - \frac{e^{\pi/2}}{2}$$.

At x = 0: $$y(0)\cdot 1 = e^0/2 - e^{\pi/2}/2 = (1-e^{\pi/2})/2$$.

The correct answer is Option (2): $$\frac{1}{2}(1-e^{\pi/2})$$.