General solution for this differential equation I want the general solution for this equation L and I'm stuck.
we have this:
$$I(y) = y , D^k(y) : = \frac{d^k}{dx^k}(y) = \frac{d^k y}{dx^k}  $$
the equation is :
$L(y):= \bigl(D-5I\bigr)^4\bigl(D^2+2D+5I\bigr)^3\bigl(D^2-4D+4I\bigr)^2(y) = 0$
here is what i did which i'm 99% sure that i'm wrong:
i started opening up the whole Ds and Is in the equation L and reached a thing and started solving it which it took me to this:
$(y'-5y)^4(y"+2y'+5y)^3(y"-4y'+4y)^2 = 0$
which makes each part equal to zero.
I'm wrong right?
please help me out...
 A: No, you're applying a composition of several commuting differential operators to $y$, so you get a high order differential equation, you won't get products of derivatives such as $y'$, $y''$, etc. What one does here is to use that the differential equation was already factored in terms of $D$ to easily solve the associated characteristic equation. The roots are:

*

*$5$, with multiplicity $4$;

*$-1\pm 2{\rm i}$, with multiplicity $3$;

*$2$, with multiplicity $4$.

So the general complex solution is $$y(t) = (a_0+a_1t+a_2t^2+a_3t^3){\rm e}^{5t} + (b_0+b_1t+b_2t^2){\rm e}^{(-1+2{\rm i})t} + (c_0+c_1t+c_2t^2+c_3t^3){\rm e}^{2t}.$$A real basis for the solution space is $$\{ {\rm e}^{5t}, t{\rm e}^{5t}, t^2{\rm e}^{5t}, t^3{\rm e}^{5t}, {\rm e}^{-t}\cos t, t{\rm e}^{-t}\cos t, t^2{\rm e}^{-t}\cos t , {\rm e}^{-t}\sin t, t{\rm e}^{-t}\sin t, t^2{\rm e}^{-t}\sin t, {\rm e}^{2t}, t{\rm e}^{2t}, t^2{\rm e}^{2t}, t^3{\rm e}^{2t} \}.$$This has $14$ elements, which is the order of the equation: $4+3\cdot 2 + 2\cdot 2 = 14$.
