Second Order Equations with Constant Coefficients

Suppose we have the equation

y'' + y = 0

We don't really have any techniques to deal with this sort of equation, so let's try guessing.

Suppose we try a polynomial:

$$y(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 = \sum_{j=0}^{j=n}a_jx^j$$

then

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

from which we quickly conclude that we must have
y(x) = 0.

Second Order Equations with Constant Coefficients

Suppose for a moment that we were allowed polynomials of infinite length, something that looks like

$$y(x) = \sum_{j=0}^{j=\infty} a_jx^j$$

Does this help?

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

$$= \sum_{j=0}^{j=\infty} (j+2)(j+1)a_{j+2}x^j$$

Which implies (from y'' = -y)

$$a_{j+2} = -\frac{a_j}{(j+2)(j+1)}$$

We can start writing out the coefficients for the series...

$$a_2 = -\frac{a_0}{2 \cdot 1} = -\frac{a_0}{2!}; \hspace{.3in} a_4 = -\frac{a_2}{4 \cdot 3} = \frac{a_0}{4!};$$

$$a_6 = -\frac{a_4}{6 \cdot 5} = -\frac{a_0}{6!}; \quad \textrm{etc.}$$

At which point we might surmise that

$$a_{2n} = \frac{(-1)^na_0}{2n!}$$