Finding the general solution of a sixth degree differential equation

Find a differential equation whose solutions are $y_1 = e^{2x} + e^{-4x}\sin(3x)$ and $y_2 = e^{-2x} + 5e^{2x}$.

Am I supposed to assume that $y_1$ and $y_2$ can take the forms:

$y_1 = Ae^{2x} + e^{-4x}[C\cos(3x)+D\sin(3x)]$

$y_2 = Ee^{-2x} + Fe^{2x}$

I'm not sure where to go from here. Any advice?

Thank you!

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

$$ ({D}^{4}+8\,{D}^{3}+21\,{D}^{2}-32\,D-100)y(x)=0 $$

$$\Rightarrow y^{(4)}+8y^{(3)}+21y^{(2)}-32y'-100y=0\,. $$