differential equation when t tends to infinity 
I think B and D are both correct. But there is supposed to be only one correct answer. Any help would be appreciated! Thank you!
 A: The trivial solution $y=0$ is one particular solution, and it doesn't satisfy D.
A: Consider the equation $x^2+bx+c$. Let the roots be $\xi_1,\xi_2$. Either both roots are negative (the sum is $-b<0$ and the product is $c>0$), and possibly equal, either they are complex conjugate with real part $-b/2<0$.
If $\xi_1\ne \xi_2$ you have two linearly independent solutions $y_i(t)=\exp(\xi_i t)$, which both satisfy B. If $\xi_1= \xi_2$ (which is then real and negative), then you have two linearly independent solutions $y_1(t)=\exp(\xi_1 t)$ and $y_2(t)=t\exp(\xi_1 t)$ which also satisfy B.
The general solution is $y(t)=k_1y_1(t)+k_2y_2(t)$, which still satisfies B. That is, whatever the solution of the differential equation, B is true.
Additionally, if $k_1=k_2=0$, then the solution is identically zero, thus it's periodic, and A is also true, so you have A,B and C. And D is true if $\xi_1$ and $\xi_2$ are complex conjugate and if $k_1$ and $k_2$ are not both zero.
Hence, if you know only one answer is correct, then it must be B, and you know also that the solution is not identically $0$, and the roots $\xi_i$ are both real and negative.
