Solve this differential equation $x^2y''-5xy'+6y=0$ Solve this equation
$$
\begin{cases}
x^2y''-5xy'+6y=0 \\
y(-1)=3 \\
y'(-1)=2
\end{cases}
$$
I got
$$y=c_1x^{3+\sqrt3}+c_2x^{3-\sqrt3}$$
I have three little questions.


*

*Could I solve the problem by substituting $(-1)^{\sqrt3}=\cos((\sqrt 3) \pi)+i\sin((\sqrt 3)\pi)$?

*I need to substitute $t=-x$?

*If i have to use 2. , isn't the problem wrong because the problem itself contains $y(-1)=3$?

 A: This is a Cauchy-Euler ODE. In order to extend your solution for $x<0$ replace $x$ by $|x|$ (or from the beginning use the substitution $t=\ln|x|$). So we have 
$$y(x)=c_1|x|^{3+\sqrt3}+c_2|x|^{3-\sqrt3}.$$
Now solve
$$\begin{cases}
y(-1)=c_1|-1|^{3+\sqrt3}+c_2|-1|^{3-\sqrt3}=3 \\
y'(-1)=-(3+\sqrt{3})c_1|-1|^{2+\sqrt3}-(3-\sqrt{3})c_2|-1|^{2-\sqrt3}=2
\end{cases}$$
and find $c_1,c_2$:
$$c_1= \frac{3-3\sqrt{3}}{2},\quad c_2 = \frac{3+3\sqrt{3}}{2}.$$
A: Well, first of all let $t=\ln(x)$, which gives $x=e^t$. Now, by the chain rule $\frac{\text{d}y(x)}{\text{d}x}=\frac{\text{d}t}{\text{d}x}\frac{\text{d}y(t)}{\text{d}t}$, and then $\frac{\text{d}^2y(x)}{\text{d}x^2}=\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y(t)}{\text{d}t}\frac{\text{d}t}{\text{d}x}\right)=\frac{\text{d}^2t}{\text{d}x^2}\frac{\text{d}y(t)}{\text{d}t}+\left(\frac{\text{d}t}{\text{d}x}\right)^2\frac{\text{d}^2y(t)}{\text{d}t^2}$. Therefore $\frac{\text{d}y(x)}{\text{d}x}=e^{-t}$ and $\frac{\text{d}^2y(x)}{\text{d}x^2}=e^{-2t}\frac{\text{d}^2y(t)}{\text{d}t^2}-e^{-2t}\frac{\text{d}y(t)}{\text{d}t}$. When you substitute this into your DE, you end up with:
$$\frac{\text{d}^2y(t)}{\text{d}t^2}-6\frac{\text{d}y(t)}{\text{d}t}+6y(t)=0\tag1$$
Now, assume a solution will be proportional to $e^{\lambda t}$ for some constant $\lambda$. Substitute $y(t)=e^{\lambda t}$ into the differential equation.
A: Others have already posted methods involving clever substitutions, but I wanted to present another, perhaps more intuitive method to solve this O.D.E. Let's write the equation again: $$x^2y''-5xy'+6y=0$$
You might notice that the $y''$ term is multiplied by a polynomial of degree $2$ in $x$, the $y'$ term a degree $1$ polynomial, and the $y$ term a degree zero polynomial. Given this structure, a clever guess solution for this ODE is $y=x^\alpha$. For the entire thing to be $=0 ~ \forall x$, we would want the degrees of all the terms to be the same. Notice that our guess solution fits that criterion perfectly: $$x^2\alpha(\alpha-1)x^{\alpha-2}-5x\alpha x^{\alpha-1}+6x^\alpha=0$$
$$\alpha(\alpha-1)x^{\alpha}-5\alpha x^{\alpha}+6x^\alpha=0$$
Cancelling the $x^\alpha$, We find $$\alpha(\alpha-1)-5\alpha +6=0$$
$$\alpha^2-6\alpha+6=0$$
$$\alpha_1=3-\sqrt{3}, \ \alpha_2=3+\sqrt{3}$$
So we have found 2 linearly independent solutions to our second order ODE and therefore our general solution may be written as $$y(x)=c_1x^{\alpha_1}+c_2x^{\alpha_2}.$$
Perhaps this was the method you used anyway but I found it a bit surprising nobody else had posted something like this yet and I thought it was worth writing.
