Extraordinary Differential Equations #3

Let $f(x) = kx + m + \cos (nx)$ . Therefore, we have:

$\begin{aligned} y' + y f(x) & = \frac {f(x)}y \\ \frac {dy}{dx} & = \left(\frac 1y - y \right)f(x) \\ \int \frac {y}{1-y^2} dy & = \int f(x) dx \\ - \frac 12 \ln (|1-y^2|) & = \frac {kx^2}2 + mx + n \sin (nx) + \color{#3D99F6} C & \small \color{#3D99F6} C \text{ is the constant of integration.} \\ - \ln (|1-y^2|) & = kx^2 + 2mx + 2n \sin (nx) + C & \small \color{#3D99F6} \text{Given that }y(0) = \sqrt 2 \\ - \ln (y^2-1) & = kx^2 + 2mx + 2n \sin (nx) + C \\ \implies 0 & = C & \small \color{#3D99F6} \text{Given that }y(-2 \pi) = \sqrt 2 \\ 0 & = 4k \pi^2 - 4m \pi \\ \implies m & = k \pi & \small \color{#3D99F6} \text{Given that }y(-\pi) = \sqrt {e+1} \\ - \ln (e+1-1) & = k\pi^2 - 2k\pi^2 \\ - 1& = - k\pi^2 \\ \implies k & = \frac 1{\pi^2} \\ - \ln (y^2-1) & = \frac {x^2}{\pi^2} + \frac {2x}\pi + 2n \sin (nx) & \small \color{#3D99F6} \text{Putting } x=\phi \pi \\ - \ln (y(\phi \pi)^2-1) & = \phi^2 + 2\phi \\ y(\phi \pi)^2-1 & = e^{-\phi (\phi + 2)} \\ y(\phi \pi) & = \sqrt{e^{-\phi (\phi + 2)}+1} & \small \color{#3D99F6} \text{Note that }y > 0 \end{aligned}$

Therefore, the answer is $\boxed{D}$ .