Finding General Solutions to 2nd Order Differential Equations, Am I on the right track? I'm studying Differential Equations, and I'm working on figuring out how to solve using Undetermined Coefficients.  We've been assured that on the test we will only have to solve equations with a 2nd order at most.  That being said, I'm trying to figure out what my general solutions should look like for each possible combination of roots.  
I've read through a lot of very verbose generalized general solutions, that I'm having a lot of trouble really comprehending.  Here's what I've taken from them.
Two Real Roots (non-repeating)
$$
Y_c = c_1e^{m_1x}+c_2e^{m_2x}
$$
Repeated Real Roots
$$
Y_c = c_1e^{x}+c_2xe^{x}
$$
Imaginary Roots (I'm really lost on this one)
$$
Y_c = c_1\cos+c_2\sin
$$
My Questions is this: Am I on the right track with the first two? What am I doing wrong with the last one?
*[We're using Dennis G Zill's A First Course in Differential Equations]*
 A: Yes you are fine on the first cases. 
For the second, a small correction. If the root if $m_1$ (it's a double root), then 
$$Y_c=c_1e^{m_1 x}+c_2xe^{m_1 x}.$$
(You left out the $m_1$ for the root.)
As for the third:
When you have a complex conjugate pair of roots,$\alpha\pm\beta i$: Then we get 
$$e^{(\alpha+\beta i)x}=e^{\alpha x}e^{\beta i x}=e^{\alpha x}[\cos(\beta x)+i\sin(\beta x)].$$
Now you can show that if you get a complex solution to a real, linear ODE then the real and imaginary parts of that solution are real solutions of that ODE. Thus, we get $e^{\alpha x}\cos(\beta x)$ and $e^{\alpha x}\sin(\beta x)$ as the solutions we seek. Moreover, they are linearly independent since their Wronskian is nonzero. Hence,
$$
Y_c(x)=c_1e^{\alpha x}\cos(\beta x)+c_2e^{\alpha x}\sin(\beta x).
$$
PS If you repeat the above process for $\alpha-\beta i$ the $Y_c$ you get won't be any different than the one above, except for the sign of one of the constants on the front, but those constants are arbitrary so we gain nothing new. The $Y_c$ above classifies what happens in the complex conjugate pair of roots case.
A: You're correct on the first one: if $m_1$ and $m_2$ are distinct roots, then
a general solution is $$Y_c = c_1 e^{m_1 x} + c_2 e^{m_2 x}$$ 
For the case of repeated roots $m$, however, it should be
$$ Y_c = c_1 e^{mx} + c_2 x e^{mx}$$
For complex roots $a \pm i b$, it should be
$$Y_c = c_1 e^{ax} \cos(bx) + c_2 e^{ax} \sin(bx)$$
