Question

Is x = 0an ordinary point of the differential equation $\mathbf{y}\mathbf{"}\mathbf{+}\mathbf{}\mathbf{5}\mathbf{xy}\mathbf{\text{'}}\mathbf{}\mathbf{+}\sqrt{\mathbf{xy}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$?

Short answer

x = 0 is a singular point of the given differential equation.

Step-by-step solution

To Find ordinary point of the given differential equation

Our aim is to find if x = 0 is an ordinary point of the given differential equation or not. First, let's assume we have a differential equation with the following form

$$\begin{gathered} a_1\left(x\right)y"+a_2\left(x\right)y\text{'}+a_3\left(x\right)y=0 \\ y"+\frac{a_2\left(x\right)}{a_1\left(x\right)}y\text{'}+\frac{a_3\left(x\right)}{a_2\left(x\right)}y=0\left.[\text{Dividing}\text{by}{a}_1(x)\right] \\ y"+P\left(x\right)y^{\text{'}}+Q\left(x\right)y=0\dots \left(1\right) \end{gathered}$$

Where,

$P\left(x\right)=\frac{a_2\left(x\right)}{a_1\left(x\right)}\text{and}Q\left(x\right)=\frac{a_3\left(x\right)}{a_1\left(x\right)}\dots$

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

A point $x=x_0$is said to be an ordinary point of the differential equation, if$P\left(x\right)$ and$Q\left(x\right)$ are analytic at that point, otherwise $x_0$is said to be a singular point.

The derivatives point

A point x = 0, this implies that you can expand this function around that point using

Talyor Series, which is calculated by the function's derivatives at that point. If the function has singular points, you will not be able to find the function's derivatives at that point, and it will blow up. Therefore, if the functions $P\left(x\right)$and $Q\left(x\right)$are infinitely differentiable at$x_0$, this implies that $P\left(x\right)$and $Q\left(x\right)$are analytic at that point and $x_0$is an ordinary point. For the given differential equation, we have

$P\left(x\right)=5xandQ\left(x\right)=\sqrt{x}=x^{-1/2}$

Now, we find the first derivative of$P\left(x\right)$at x = 0.

$$\begin{gathered} \frac{d}{dx}\left[Q\left(x\right)\right]=\frac{d}{dx}\left[x^{1/2}\right] \\ =\frac{1}{2}{x}^{-1/2} \\ =\frac{1}{2\sqrt{x}} \\ \underset{x\to 0}{\lim }\left[\frac{d}{dx}Q\left(x\right)\right]=\underset{x\to 0}{\lim }\frac{1}{2\sqrt{x}} \\ =doesn\text{'}texist \end{gathered}$$

Final Answer

Therefore,$Q\left(x\right)$ is not analytic at x = 0 , which implies that x = 0 is a singular point of the given differential equation.

x = 0 is a singular point of the given differential equation.