Imaginary differential equation!

The given differential equation is the first-order ODE of the form $\dfrac {dy}{dx} + p(x)y = g(x)$ . We can use a multiplier $\mu (x) = e^{\int p(x)\ dx} = e^{\int i \ dx} = e^{ix}$ . Then we have:

$\begin{aligned} e^{ix} \frac {dy}{dx} + i e^{ix} y & = 2 e^{ix} \color{#3D99F6} \sin x & \small \color{#3D99F6} \text{By Euler's formula: } e^{i\theta} = \cos \theta + i\sin \theta \\ \frac d{dx} \left(e^{ix} y\right) & = 2 e^{ix} \color{#3D99F6} \left(\frac {e^{ix}-e^{-ix}}{2i}\right) \\ & = i \left(1 - e^{2ix}\right) \\ \implies e^{ix}y & = \int i \left(1 - e^{2ix}\right) dx \\ & = ix - \frac {e^{2ix}}2 + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration,} \\ \implies y(x) & = ix e^{-ix} - \frac {e^{ix}}2 + Ce^{-ix} \\ y(0) & = 0 - \frac 12 + C = \frac 32 & \small \color{#3D99F6} \implies C = 2 \\ \implies y(x) & = ix e^{-ix} - \frac {e^{ix}}2 + 2e^{-ix} \\ y(\pi) & = i \pi e^{-\pi i} - \frac {e^{\pi i}}2 + 2e^{-\pi i} & \small \color{#3D99F6} \text{By Euler's formula: } e^{\pm \pi i} = - 1 \\ & = - \pi i + \frac 12 - 2 \\ & = - \frac 32 - \pi i \end{aligned}$

Therefore, $abc = \boxed 6$ .