How to solve this Euler-Cauchy equation $2x^2y''-xy'+6y=0$
$2r^2-3r+6$ Is this my characteristic equation? 
Now, I know that I should get a complex answer, and then use Euler's identity to simplify it into sines and cosines. However, I really can't find a great walkthrough online, and so I'm wondering if anyone could give me a thorough explanation on how to do these. I have a lot of them to do, so I'd appreciate an example! Thanks! 
 A: Write your equation as $$2x^2\frac{{\rm d}^2y}{{\rm d}x^2}-x\frac{{\rm d}y}{{\rm d}x}+6y = 0.$$Do the substituiton $x = e^t$. We have: $$t = \ln x \implies \frac{{\rm d}t}{{\rm d}x} = \frac{1}{x} = \frac{1}{e^t} = e^{-t}.$$
By the chain rule: $$\frac{{\rm d}y}{{\rm d}x} = \frac{{\rm d}y}{{\rm d}t}\frac{{\rm d}t}{{\rm d}x} = e^{-t}\frac{{\rm d}y}{{\rm d}t},$$ and: $$\begin{align} \frac{{\rm d}^2y}{{\rm d}x^2} &= \frac{{\rm d}}{{\rm d}x}\left(e^{-t}\frac{{\rm d}y}{{\rm d}t}\right) \\ &= \frac{{\rm d}}{{\rm d}x}(e^{-t})\frac{{\rm d}y}{{\rm d}t}  +e^{-t} \frac{{\rm d}}{{\rm d}x}\left(\frac{{\rm d}y}{{\rm d}t}\right)  \\  &= \frac{{\rm d}}{{\rm d}t}(e^{-t})\frac{{\rm d}t}{{\rm d}x}\frac{{\rm d}y}{{\rm d}t}+e^{-t}\frac{{\rm d}}{{\rm d}t}\left(\frac{{\rm d}y}{{\rm d}t}\right)\frac{{\rm d}t}{{\rm d}x} \\ &= -e^{-2t}\frac{{\rm d}y}{{\rm d}t}+e^{-2t}\frac{{\rm d}^2y}{{\rm d}t^2} \\ &= e^{-2t}\left(\frac{{\rm d}^2y}{{\rm d}t^2} - \frac{{\rm d}y}{{\rm d}t}\right).\end{align}$$
Plugging this into your equation yields: $$\begin{align} 2e^{2t}e^{-2t}\left(\frac{{\rm d}^2y}{{\rm d}t^2} - \frac{{\rm d}y}{{\rm d}t}\right)-e^te^{-t}\frac{{\rm d}y}{{\rm d}t}+6y &= 0 \\ 2\frac{{\rm d}^2y}{{\rm d}t^2}-3\frac{{\rm d}y}{{\rm d}t} +6y &= 0,\end{align}$$ which indeed has $2r^2-3r+6 = 0$ as a characteristic equation. Solving: $$r = \frac{3\pm \sqrt{9-4\cdot 2\cdot 6}}{4} = \frac{3 \pm \sqrt{39}{\rm i}}{4}.$$
So the general solution is: $$\begin{align} y &= c_1 e^{3t/4}\cos\left(\frac{\sqrt{39}}{4}t\right)+c_2e^{3t/4}\sin\left(\frac{\sqrt{39}}{4}t\right) \\ y &= c_1e^{3 (\ln x)/4}\cos\left(\frac{\sqrt{39}}{4}\ln x\right)+c_2e^{3 (\ln x)/4}\sin\left(\frac{\sqrt{39}}{4}\ln x\right) \\  y &= c_1 \sqrt[4]{x^3}\cos\left(\frac{\sqrt{39}}{4}\ln x\right)+c_2\sqrt[4]{x^3}\sin\left(\frac{\sqrt{39}}{4}\ln x\right), \quad c_1,c_2 \in \Bbb R.\end{align}$$
A: Your characteristic equation is correct. The solution is 
$$\frac{3}{4}\pm \frac{\sqrt{39}}{4} i$$
So the solution of the differential equation is
$$c_1 x^{3/4}\cos{(\frac{\sqrt{39}}{4}\ln{x})}+c_2 x^{3/4}\sin{(\frac{\sqrt{39}}{4}\ln{x})}$$
In general, if the solution of characteristic equation are two real numbers $r_1,r_2$, then the solution of the DE is 
$$c_1 x^{r_1}+c_2 x^{r_2}$$
If the solution of the characteristic equation is one real number $r$, the solution of the DE is
$$c_1 x^r \ln{x} +c_2 x^r$$
If the solution of characteristic equation are complex numbers $\alpha\pm \beta i$, then
$$c_1 x^{\alpha}\cos{(\beta \ln{x})}+c_2 x^{\alpha}\sin{(\beta \ln{x})}$$
