Solving it by writing out a characteristic equation is not helping me find the solution to the above equation. Any help would be appreciated thanks.


One way to do it is by expanding $y(x)$ as a series:

$$ y(x) = \sum_{n=0}^\infty c_n x^n $$

We can then write $y''(x)$ as:

$$ y''(x) = \sum_{n=2}^\infty c_n n (n-1) x^{n-2} $$

and $x^2 y(x)$ as:

$$ x^2 y(x) = \sum_{n=0}^\infty c_n x^{n+2} $$

Now we re-parameterize the sums so they both have $x^n$ in the summation:

$$ y''(x) = \sum_{n=0}^\infty c_{n+2} (n+2) (n+1) x^n $$

$$ x^2 y(x) = \sum_{n=2}^\infty c_{n-2} x^n $$

Finally we equate the coefficients (while being careful about the limits of the sums):

$$ 2 c_2 = 0 $$

$$ 6 c_3 = 0 $$

$$ c_{n+2} (n+2) (n+1) = c_{n-2} \; \forall n \geq 2 $$

We can solve this by rewriting the last line as

$$ c_{n+4} = \frac{c_n}{(n+4)(n+3)} \; \forall n $$

and noticing that we can come up with two clear independent solutions: one where $c_0 = 1$ and $c_1 = 0$, and vice versa. In either case, for any $n \equiv 2\mod 4$ or $n \equiv 3\mod 4$, we have $c_n = 0$.

The first solution becomes:

$$ y(x) = 1 + \frac{1}{12} x^4 + \frac{1}{672} x^8 + \cdots $$

and the second solution is:

$$ y(x) = x + \frac{1}{20} x^5 + \frac{1}{1440} x^9 + \cdots $$

  • $\begingroup$ This is fine but I wonder if we are not missing a complex part as Dmoreno showed in his answer which can also write $$c_1 e^{-\frac{x^2}{2}} H_{-\frac{1}{2}}(x)+c_2 e^{\frac{x^2}{2}} H_{-\frac{1}{2}}(i x)$$ where appear Hermite polynomials. $\endgroup$ May 25 '15 at 4:57
  • 1
    $\begingroup$ @ClaudeLeibovici If you want to involve complex values, that can be done by superpositions of the two independent solutions with complex coefficients. These solutions are linearly independent and satisfy the differential equation (which is a linear second order equation), so all solutions can be represented in terms of them. $\endgroup$ May 25 '15 at 5:03
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    $\begingroup$ As I said, your answer is nice and I have no problem at all with your comments. It is probably the first time I see problem like this one where the closed form solutions contain such mixtures ! Cheers :-) $\endgroup$ May 25 '15 at 5:13

The solutions of this differential equations are well known and can be written in terms of the Parabolic Cylinder functions. In particular, it can be found that:

$$y(x) = A \, D_{-1/2} (\mathrm{i} \sqrt{2} x ) + B \, D_{-1/2} ( \sqrt{2} x ) $$ is the general solution of your equation.

Hope this helps!

  • $\begingroup$ This is not correct. Airy functions are solutions to $y''(x) = x y(x)$, not $y''(x) = x^2 y(x)$. $\endgroup$ May 25 '15 at 1:40
  • $\begingroup$ i thought that was the solution to $y'' = xy.$ $\endgroup$
    – abel
    May 25 '15 at 1:41
  • $\begingroup$ Apologies. I misread the question! $\endgroup$
    – Dmoreno
    May 25 '15 at 1:41

Try factoring

$$ \left(\frac{d^2}{dx^2} - x^2\right) = \left(\frac{d}{dx}-x\right)\left(\frac{d}{dx}+x\right)-1 $$ I suspect that solutions can be found simliarly to how it is done for the quantum harmonic oscillator.

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    $\begingroup$ May I ask why is there a 1 there? $\endgroup$
    – Chee Han
    May 25 '15 at 1:54
  • $\begingroup$ I tried this way but I could'nt figure out how to go on from here $\endgroup$ May 25 '15 at 2:25
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    $\begingroup$ @CheeHan Because $\frac{d}{dx}$ and $x$ do not commute (they are operators) $\endgroup$ May 25 '15 at 2:34
  • $\begingroup$ I just found out a mistake on my expansion, thanks! $\endgroup$
    – Chee Han
    May 25 '15 at 3:00

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