Series Solutions: Hermite's Equation

Exercise 3:

Consider the Hermite Equation of order 5:


Find the solution satisfying the initial conditions a0=1, a1=0.


The solution will not be a polynomial.

Since a1=0, all odd coefficients will be zero. Let's compute a few of the even coefficients:

\begin{eqnarray*}a_2&=&\frac{2(0-5)}{1\cdot 2}\\
a_4&=&\frac{2(2-5)}{3\cdot 4}a...
a_6&=&\frac{2(4-5)}{5\cdot 6} a-4=\frac{2^3 (-5)(-3)(-1)}{6!}

From this it is not too hard to come up with the general formula:

\begin{displaymath}a_{2n}=\frac{2^n(-5)(-3)(-1)\cdots(2n-7)}{(2n)!}\mbox{ for all }n=0,1,2,\ldots\end{displaymath}

This leads to the formula

\begin{displaymath}y(t)=\sum_{n=0}^\infty \frac{2^n(-5)(-3)(-1)\cdots(2n-7)}{(2n)!} t^{2n}\end{displaymath}

for the solution to the given initial value problem.

The solution converges and solves the equation for all real numbers.

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