Question

In Problems 19 and 200 investigate whether x = 0 is an ordinary point, singular point, or irregular singular point of the given differential equation. [Hint: Recall the Maclaurin series for cos x and ex .]

$xy"+(1-\cos x)y\text{'}+x^{2}y=0$

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Short answer

The point x = 0 is an ordinary point of the given differential equation.

Step-by-step solution

Step 1: Definition of Power series solutions

If $X=X_0$ is an ordinary point of the differential equation , we can always find two linearly independent solutions in the form of a power series centered at X₀, that is,

$y=\sum _{n=0}^x{C}_n{(X-X_0)}^n.$

A power series solution converges at least on some interval defined by$\left|X-X_0\right|<R$, where R is the distance from$X_0$to the closest singular point.

Using Maclaurin series to solve the problem

We first convert the given differential equation in standard form to get

$y"+\left(\frac{1-\cos x}{x}\right)y\text{'}+xy=0$

Now, we compare the above equation to the standard DE form

$y+P\left(x\right)y+Q\left(x\right)y=0$

which gives

$P\left(x\right)=\frac{1-\cos x}{x}andQ\left(x\right)=X$

Now, we recall that the Maclaurin series for Cos X is

$Cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$

Substituting the value offrom above inthen gives

$P\left(x\right)=\frac{1-\cos x}{x}$

$$\begin{gathered} =\frac{1-\left(1-{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^4}{4!}}-{\displaystyle \frac{x^6}{6!}}+...\right)}{x.} \\ =\frac{{\displaystyle \frac{x^2}{2!}}-{\displaystyle \frac{x^4}{4!}}-{\displaystyle \frac{x^6}{6!}}-..}{x} \\ =\frac{x}{2!}-\frac{x^3}{4!}+\frac{x^6}{6!}-... \end{gathered}$$

Since, both the coefficients P(x) and Q(x) are analytic at x=0 we can say that the point is said to be an ordinary point of the given differential equation from definition 6.2.1.

The point X =0 is an ordinary point of the given differential equation.