The general solution of the ODE $x^2y''+5xy'+13y=0$ I am trying to solve the ODE $$x^2y''+5xy'+13y=0$$.
Need a confirmation that answer is   $$y=\dfrac{1}{x^2}\bigg(c_1 \cos(3\ln x)+c_2 \sin(3\ln x)\bigg)$$
I do not understand what quality standards are on this website .
 A: Solve the auxiliary equation $$am^2+(b-a)m+c=0$$ wherein $a=1$,the coefficient of $x^2$, $b=5$,the coefficient of $x$ and $c=13$
for finding the possible $m$'s. Now, if $m_1,m_2$ are distinct solutions so the general solution of your Cauchy-Euler ODE will be as $$y_c=C_1x^{m_1}+C_2x^{m_2}$$ If you have $m_1=m_2=m=\frac{a-b}{2a}$ then $y_c=C_1x^m+C_2x^m\ln(x)$ and finally if you have $m=\alpha\pm i\beta$ then $$y_c=x^{\alpha}(C_1\cos(\beta\ln x)+C_2\sin(\beta\ln x))$$ where $C_1,C_2$  are constants. Of course $x\in(0,+\infty)$. Your solution looks fine.
A: You can substitute any values in the "arbitrary constants" to check but it would certainly be not a general solution.
The general solution of a differential equation represent a family of curves , satisfying the above equation.So, by substitution you pick up any arbitrary curve from the family and check the differential equation.
Or i'll suggest just construct the DE from the equation you got. differentiate it twice and eliminate the constants from the equations you get . If the DE you get is same , then the answer is correct. 
