This is a follow Up question To roots of a linear DE $(D^2+2cD+k)y=0$
Let y(x) be a non-trivial solution of the second order linear differential equation
$$\dfrac{d^2y}{dx^2}+2k\dfrac{dy}{dx}+ly=0 $$
MY approach The charecteristic equation is $m^{2}+2km+l=0$$\Longrightarrow m=-k\pm\sqrt{k^{2}-l}$
In Option (C) If $k^{2}-l<0$ $\Longrightarrow$Solution is y=$e^{-kx}\left(c_{1}cosx\left(\sqrt{l-k^{2}}\right)+c_{2}sinx\left(\sqrt{l-k^{2}}\right)\right)$$\Longrightarrow$ y$\longrightarrow0$ as x$\longrightarrow\infty$
If $k^{2}-l=0$ $\Longrightarrow$Solution is y=$\left(c_{1}+c_{2}x\right)e^{-kx}$$\Longrightarrow$y$\longrightarrow0$ as x$\longrightarrow\infty$
Note Some notations are changed .Do not get confuse with that ,See the question in Picture.
you must do case work: $$k^2-l<0$$ or $$k^2-l=0$$ or $$k^2-l>0$$ and the solution is given by $$y \left( x \right) ={\it \_C1}\,{{\rm e}^{ \left( -k+\sqrt {{k}^{2}-l} \right) x}}+{\it \_C2}\,{{\rm e}^{ \left( -k-\sqrt {{k}^{2}-l} \right) x}} $$