verification of solution of differential equation Verify that $$y= (c_1(x^{-1})) + (c_2(x^5))$$ is a solution of $$(x^2y'') - (3xy) - 5y =0$$ on any interval $[a,b]$ that does not contain an origin. If $x$ is not equal to $0$ and if $y_1$ and $y_2$ are arbitrary, show directly that $c_1$ and $c_2$ can be chosen in one and only one way so that the value of $y$ at $x$ is equal to $y_1$ and value of first differential of $y$ at $x$ is equal to $c_2$.
I tried putting the value of $y$ into equation, but the answer is not coming out equal to $0$. I am not able to verify the first part itself. It is not coming out equal to $0$. 
Please help.
 A: Although, the @AWertheim's approach is a routine way; there is another approach for this OE and your question. As you may know any ODE like $$ax^2y''+bxy'+cy=0$$ in which $a,b,~c$ are constants is said to be a Cauchy-Euler equation. It is homogeneous and so according to its auxiliary equation $$am^2+(b-a)m+c=0$$ it will have a general solution as $$y_c=C_1y_1(x)+C_2y_2(x)$$ Here we have $a=1, ~b=-3,c=-5$. So $m^2-4m-5=0$ or $(m+1)(m-5)=0$ and so $$m_1=-1,~~m_2=5$$ which leads us to have $$y_c=C_1x^{m_1}+C_2x^{m_2}=C_1x^{-1}+C_2x^{5}$$
A: I think you have a typo in your post, or a typo in the problem, and the term labeled $3*x*y$ should be $3*x*y'$. If this is indeed the case, see this post.
Computing $y''$:
$$y' = -C_{1}x^{-2} + 5C_{2}x^{4}$$
Differentiating again, we find:
$$y'' = 2C_{1}x^{-3} + 20C_{2}x^{3}$$
Now, plugging in:
$$x^{2}(2C_{1}x^{-3} + 20C_{2}x^{3}) - 3*x*(-C_{1}x^{-2} + 5C_{2}x^{4}) - 5*(C_{1}x^{-1} + C_{2}x^{5}) = 2C_{1}x^{-1} + 20C_{2}x^{5} + 3C_{1}x^{-1} - 15C_{2}x^{5} - 5C_{1}x^{-1} - 5C_{2}x^{5} = 0$$
So $y$ is indeed a solution on any closed interval not containing the origin (where $y$ and its derivatives are discontinuous). 
Can you try to solve the second part of the problem now? I'll give you a hint. The solution $y$ is a linear combination of two linearly independent terms (if you feel compelled to verify this, take the Wronskian of $x^{-1}$ and $x^{5}$). Now, imposing two initial conditions on $y$ by requiring $y(x) = y_{1}$ and $y'(x) = y_{2}$ gives you a linear system of equations. Assuming that the system has full rank (which shouldn't be hard to show), then the solution must be unique. 
