Question

Is x = 0 an ordinary or a singular point of the differential equation$\mathbf{xy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{\left(}\mathbf{sinx}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{0}$? Defend your answer with sound mathematics. [Hint: Use the Maclaurin series of$\mathbf{sinx}$and then examine$\mathbf{\left(}\mathbf{sinx}\mathbf{\right)}\mathbf{/}\mathbf{x}$.]

Short answer

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

Step-by-step solution

To Find a singular point or an ordinary point of the given differential equation.

We have;

$xy\text{'}\text{'}+\left(sinx\right)y=0$

We aim to find if x = 0 is a singular point or an ordinary point of the given differential equation. First, let's assume we have a differential equation with the following form;

$$\begin{gathered} a_1\left(x\right)y\text{'}\text{'}+a_2\left(x\right)y\text{'}+a_3\left(x\right)y=0 \\ y\text{'}\text{'}+\frac{a_2\left(x\right)}{a_1\left(x\right)}y\text{'}+\frac{a_3\left(x\right)}{a_1\left(x\right)}y=0 \left[Dividing by a_1\left(x\right)\right] \\ y\text{'}\text{'}+P\left(x\right)y\text{'}+Q\left(x\right)y=0 \dots \left(1\right) \end{gathered}$$

Where,

localid="1664859187547" $\mathrm{P}\left(\mathrm{x}\right)=\frac{{\mathrm{a}}_2\left(\mathrm{x}\right)}{{\mathrm{a}}_1\left(\mathrm{x}\right)}\mathrm{and}\mathrm{Q}\left(\mathrm{x}\right)=\frac{{\mathrm{a}}_3\left(\mathrm{x}\right)}{{\mathrm{a}}_1\left(\mathrm{x}\right)}\dots$

The ordinary point of the differential equation.

Equation (1) is the standard form, upon which we will build our analysis. We recall the following definition.

A point $\mathit{x}\mathbf{=}{\mathit{x}}_{\mathbf{0}}$is said to be an ordinary point of the differential equation, if$\mathit{P}\left(x\right)$and$\mathit{Q}\mathbf{\left(}\mathit{x}\mathbf{\right)}$are analytic at that point, otherwise${\mathit{x}}_{\mathbf{0}}$is said to be a singular point.

If you have an analytic function at a point, this implies that you can expand this function around that point using Taylor series, which is calculated by the function's derivatives at that point.

If the function has singular point, you will not be able to find the function's derivatives at that point, and it will blow up. Investigating (2), you will be able to deduce that $\underset{¯}{P\left(x\right)}$and $Q\left(x\right)$are not analytic at the points, which will make the denominator equal to zero, after common factors reduction. Now, we put the given differential equation into the standard form.

$$\begin{gathered} \frac{x}{x}y"+\frac{\left(sinx\right)}{x}y=0 [Dividingby x] \\ y"+\frac{\left(sinx\right)}{x}y=0 \end{gathered}$$

On comparing with standard form, we get:

localid="1664859199903" $\mathrm{P}\left(\mathrm{x}\right)=0\mathrm{and}\mathrm{Q}\left(\mathrm{x}\right)=\frac{\left(\mathrm{sinx}\right)}{\mathrm{x}}$

Final Answer.

Now, using the Maclaurin series of the sine function;

$$\begin{gathered} Q\left(x\right)=\frac{1}{x}\left[x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots \right] \\ =1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\dots \dots \left(4\right) \end{gathered}$$

From (3) and (4), $\underset{¯}{P\left(x\right)}$and $Q\left(x\right)$are infinitely differentiable at $x=0$, which implies that $\underset{¯}{P\left(x\right)}$and $Q\left(x\right)$are analytic at $x=0$and x = 0 is an ordinary point of the given differential equation.

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