Please help me check my solution, because I'm not sure that it is correct or not.

$\displaystyle \frac{d^4y}{dx^4}-7\frac{d^2y}{dx^2}-18y=0$

$\displaystyle m^4e^{mx}-7m^2e^{mx}-18e^{mx} = 0$

$\displaystyle (m^4-7m^2-18)(e^{mx}) = 0$

$\displaystyle (m-3)(m+3)(m^2+2) = 0$

$\displaystyle m = 3, -3, \sqrt{2}i, -\sqrt{2}i$

Then, $\displaystyle y=c_1e^{3x}+c_2e^{-3x}+c_3cos(\sqrt{2}x)+c_4sin(\sqrt{2}x)+c_5cos(\sq rt{2}x)-c_6sin(\sqrt{2}x)$