are defined for \(x\) near \(x_0\text{.}\) If \(P(x_0) = 0\text{,}\) then we say \(x_0\) is a singular point. Handling singular points is harder than ordinary points and so we now focus only on ordinary points.
Let us try a power series solution near \(x_0 = 0\text{,}\) which is an ordinary point. Every point is an ordinary point in fact, as the equation is a constant-coefficient one. We already know we should obtain exponentials or the hyperbolic sine and cosine, but let us pretend we do not know this fact.
The equation above is called a recurrence relation for the coefficients of the power series. It does not matter what \(a_0\) or \(a_1\) are. They can be arbitrary. But once we pick \(a_0\) and \(a_1\text{,}\) all other coefficients are determined by the recurrence relation.
Of course, in general we will not be able to recognize the series that appears, since usually there will not be any elementary function that matches it. In that case, we will be content with the series.
We jump in steps of three. First, since \(a_2 = 0\text{,}\) we must have \(a_5 = 0\text{,}\)\(a_8 = 0\text{,}\)\(a_{11}=0\text{,}\) etc. In general, \(a_{3n+2} = 0\text{.}\)
and write the general solution to the equation as \(y(x)= a_0 y_1(x) + a_1 y_2(x)\text{.}\) If we plug in \(x=0\) into the power series for \(y_1\) and \(y_2\text{,}\) we find \(y_1(0) = 1\) and \(y_2(0) = 0\text{.}\) Similarly, \(y_1'(0) = 0\) and \(y_2'(0) = 1\text{.}\) Therefore \(y = a_0 y_1 + a_1 y_2\) is a solution that satisfies the initial conditions \(y(0) = a_0\) and \(y'(0) = a_1\text{.}\)
The functions \(y_1\) and \(y_2\) cannot be written in terms of the elementary functions that you know. See Figure 7.3 for the plot of the solutions \(y_1\) and \(y_2\text{.}\) These functions have many interesting properties. For example, they are oscillatory for negative \(x\) (like solutions to \(y''+y=0\)) and for positive \(x\) they grow without bound (like solutions to \(y''-y=0\)).
This recurrence relation actually includes \(a_2 = -na_0\) (which comes about from the constant term \(2a_2+2na_0 = 0\)). Again \(a_0\) and \(a_1\) are arbitrary.
We remark that if \(n\) is a positive even integer, then \(y_1(x)\) is a polynomial as all the coefficients in the series beyond degree \(n\) are zero. If \(n\) is a positive odd integer, then \(y_2(x)\) is a polynomial. For example, if \(n=4\text{,}\) then
In the following exercises, when asked to solve an equation using power series methods, you should find the first few terms of the series, and if possible find a general formula for the \(k^{\text{th}}\) coefficient.
The methods work for other orders than second order. Try the methods of this section to solve the first-order system \(y'-xy = 0\) at the point \(x_0 = 0\text{.}\)
Attempt to solve \(x^2 y'' - y = 0\) at \(x_0 = 0\) using the power series method of this section (\(x_0\) is a singular point). Can you find at least one solution? Can you find more than one solution?