You can use the annihilator method. For example, the function $ {\rm e}^{\alpha x} $ has the annihilator $D-\alpha $, where $D=\frac{d}{dx}$. That means
$ (D-\alpha){\rm e}^{\alpha x}= \alpha {\rm e}^{\alpha x}-\alpha {\rm e}^{\alpha x} = 0 \,. $
Now, since $y_1$ and $y_2$ are solutions, then $y_1+y_2$ is a solution too.
$ y(x) = 6e^{2x} + e^{-4x} \sin(3x) + e^{-2x}= 6e^{2x}+ \frac{1}{2i}e^{(-4+3i)x}- \frac{1}{2i}e^{(-4-3i)x} + e^{-2x}\,. $
To annihilate the above equation, we apply the above annihilators to both sides of the equation
$ (D+2)(D-(-4-3i))(D-(-4+3i))(D-2)y(x) = 0 \,. $
Multiplying and simplifying the left hand side gives a differential equation of fourth order
$ ({D}^{4}+8\,{D}^{3}+21\,{D}^{2}-32\,D-100)y(x)=0 $
$\Rightarrow y^{(4)}+8y^{(3)}+21y^{(2)}-32y'-100y=0\,. $