Question

Find all singular points of the given equation and determine whether each one is regular or irregular. \(x y^{\prime \prime}+e^{x} y^{\prime}+(3 \cos x) y=0\)

Short answer

Question: Identify and classify the singular points for the given differential equation \(x y^{\prime\prime} + e^x y^\prime + (3 \cos x)y = 0\). Answer: The given differential equation has one singular point, \(x=0\), and it is a regular singular point.

Step-by-step solution

Identify the singular points

For the given equation, \(x y^{\prime\prime} + e^x y^\prime + (3 \cos x)y = 0\), the coefficient of the highest derivative (\(y^{\prime\prime}\)) is \(x\). A singular point occurs when the coefficient of the highest derivative is zero: $$x = 0$$ So, we have one singular point \(x = 0\).

Check whether the singular point is regular or irregular

We need to check if the coefficients of lower-order derivatives, when divided by the highest derivative's coefficient, have a power series expansion about the singular point. Let's look at the ratio of the coefficients of the first and second derivatives: $$\frac{e^x}{x} \quad \text{and} \quad \frac{3\cos x}{x}$$ The ratio of the first derivative to its coefficient, \(\frac{e^x}{x}\), does not have any issues near \(x=0\), and we can use the L'Hôpital's rule to show that: $$\lim_{x \to 0} \frac{e^x}{x} = \lim_{x \to 0} \frac{e^x}{1} = 1$$ However, the ratio of the second derivative to its coefficient, \(\frac{3\cos x}{x}\), is not well-defined when \(x = 0\). We can investigate the limit of this ratio as \(x\) approaches 0: $$\lim_{x \to 0} \frac{3\cos x}{x}$$ Using L'Hôpital's rule, we get: $$\lim_{x \to 0} \frac{-3\sin x}{1} = 0$$ Now that we have found the limits of both ratios, we see that they both have a finite limit when \(x\) approaches the singular point. Therefore, the singular point \(x = 0\) is a regular singular point. In conclusion, for the given differential equation \(x y^{\prime\prime} + e^x y^\prime + (3 \cos x)y = 0\), there is only one singular point, \(x=0\), and it is a regular singular point.

Key concepts

Regular Singular Point

Understanding the nature of singular points in differential equations is pivotal for predicting the behavior of their solutions. A singular point is where the coefficient of the highest-order derivative in a differential equation becomes zero, which can complicate finding solutions. But not all singular points are created equal: they can be classified as either regular or irregular.

To discern whether a singular point is regular or irregular, we examine the behavior of the coefficients of lower derivatives when divided by the coefficient of the highest derivative. More formally, a point \(x = x_0\) is a regular singular point if the functions \( (x-x_0)^n \(y^{(n)}/y^{(m)}\) \) have a power series expansion about \(x_0\) for each derivative \(y^{(n)}\) of order less than \(m\).

In our exercise, we found the singular point to be at \(x = 0\). Then we proceeded to check the ratios of the coefficients of the lower derivatives to the highest one, \(y^{\text{\prime\prime}}\). If these ratios can be expressed as a power series around \(x = 0\), or if their limits as \(x\) approaches 0 exist and are finite, we can conclude that \(x = 0\) is a regular singular point. Utilizing this approach in our example, it's confirmed that the singular point of \(x = 0\) for the given equation is indeed regular.

L'Hôpital's Rule

Encountering indeterminate forms, such as \(0/0\) or \(\infty/\infty\), when limiting a function, is where L'Hôpital's rule comes in handy. This powerful mathematical tool resolves these ambiguous limits by comparing the rates of change (derivatives) of the numerator and denominator.

According to L'Hôpital's Rule, if we have a limit \(\lim_{x \to c} \frac{f(x)}{g(x)}\) that evaluates to an indeterminate form, we can find the limit of the derivatives instead: \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\). By applying this rule one or more times, we can often simplify the expression enough to determine the limit.

In the context of our example with the ratio \(\frac{e^x}{x}\), L'Hôpital's rule allows us to resolve the indeterminate form at \(x = 0\) by finding the derivatives of the numerator and the denominator. We find that \(\lim_{x \to 0} \frac{e^x}{x} = 1\), confirming that the limit is finite and helping us to conclude that the singular point is regular.

Power Series Expansion

A power series expansion is a representation of a function as an infinite sum of terms, each of which is a power of the variable multiplied by a coefficient. It's essentially expressing a function as a polynomial of infinite degree, one that can be as simple as the Taylor series or Maclaurin series.

The power series expansion can take the form \( f(x) = \sum_{n=0}^{\infty} a_n(x-c)^n \) for constants \( a_n \) and around a center \( c \). If a function is analytic at point \( c \) (it could be represented by a power series around \( c \) and converges to the function within some interval), then this expansion can provide valuable approximation to the function's behavior near \(c\).

Returning to our exercise, the examination of whether the ratios of the differential equation coefficients could be expressed as a power series was crucial. We needed to ensure that these coefficients when expanded in a series, would have a specific behavior at \( x = 0 \) to determine the nature of the singular point. Happily, in this instance, they did conform to such a series, granting us further proof that our singular point is regular.