# Second-order Euler Cauchy; regarding choice of $\lambda$

The exact equation is: $$\quad y'' - \dfrac{4}{x}y' + \dfrac{6}{x^2}y = 0, \quad x > 0.$$

So, solving it was okay, but I literally always stumble upon the same dilemma. What do I choose to be $\lambda_1$, and what do I choose to be $\lambda_2$? Is there some golden rule I missed that you always choose the value closest to zero?

Annoys me to get the solution: $$y_{h}(x) = c_1 x^3 + c_2 x^2$$

Instead of the opposite and correct solution at both Wolfram Alpha and the solution set: $$y_h (x) = c_1 x^2 + c_2 x^3$$

So is there really a difference? Are they equivalent? This might be but a small nuisance..!

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The two expressions are identical. $c_1$ and $c_2$ could be fixed by some boundary conditions, say $y(1)=5$ and $y'(1)=14$ will give $c_1=1$ and $c_2=4$ for WA writing. – Claude Leibovici Apr 29 '14 at 12:24
I'm aware of the latter, Claude! I realise my question was pretty dumb after-all. What I was really asking is more of the type Java code conventions, i.e. if maths in this case has a convention of selecting arbitrary values and their paired answers in a specific way. :) – Erlend Apr 29 '14 at 12:27