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!
 A: 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
$$ ({D}^{4}+8\,{D}^{3}+21\,{D}^{2}-32\,D-100)y(x)=0 $$ 
$$\Rightarrow y^{(4)}+8y^{(3)}+21y^{(2)}-32y'-100y=0\,. $$  
A: $$y_1=e^{2x}+\frac12\left(e^{(-4+3i)x}-e^{(-4-3i)x}\right)$$
Now take a differential equation of the form 
$$ay+by'+cy''+y'''=0$$
Sub in $y=e^{rx}$:
$$e^{rx}(a+br+cr^2+r^3)=0$$
So we want to find a cubic $r^3+cr^2+br+a$ that has its roots as the coefficients in the above exponents. So we chose
$$(r-2)(r+4-3i)(r+4+3i)=0$$
which will generate each exponential in the expression for $y_1$ as solutions. Expanding, we get
$$r^3+6r^2+9r-50=0$$
So that $y_1$ is a solution of
$$-50y+9y'+6y''+y'''=0$$
The case for $y_2$ proceeds similarly, with a 2nd order differential equation instead.
