Question

Find all the regular singular points of the given differential equation. Determine the indicial equation and the exponents at the singularity for each regular singular point. \(2 x(x+2) y^{\prime \prime}+y^{\prime}-x y=0\)

Short answer

Answer: The regular singular points are \(x_1 = 0\) and \(x_2 = -2\), and the exponents at the singularity are \(r_1 = 1\) and \(r_2 = -\frac{1}{2}\).

Step-by-step solution

Identify regular singular points

First, let's rewrite the given equation as: \begin{equation} y^{\prime\prime} + \frac{1}{2x(x+2)} y^{\prime} - \frac{1}{2(x+2)} y = 0 \end{equation} Now, we will analyze the two coefficients of the rational functions: \(\frac{1}{2x(x+2)}\) and \(-\frac{1}{2(x+2)}\). The first coefficient becomes infinite or undefined at \(x = 0\) and \(x = -2\). The second coefficient becomes infinite or undefined at \(x = -2\). Therefore, we have two regular singular points: \(x_1 = 0\) and \(x_2 = -2\).

Apply the Frobenius method

We will now apply the Frobenius method to the given equation to find the indicial equation and exponents for each regular singular point. To do this, we will assume a solution of the form \(y(x) = \sum_{k=0}^\infty a_k x^{k+r}\), where \(r\) is the exponent.

Derivatives of the assumed solution

Start by finding the first and second derivatives of the assumed solution: \begin{align*} y^{\prime}(x) &= \sum_{k=0}^\infty (k+r) a_k x^{k+r-1} \\ y^{\prime\prime}(x) &= \sum_{k=0}^\infty (k+r)(k+r-1) a_k x^{k+r-2} \end{align*}

Substitute derivatives into the differential equation

Now, substitute the derivatives into the equation, and factor out the coefficients of \(x^r\) from both sides: \begin{equation*} 2x(x+2)\sum_{k=0}^\infty (k+r)(k+r-1) a_kx^{k+r-2} + \sum_{k=0}^\infty (k+r) a_k x^{k+r-1} - x\sum_{k=0}^\infty a_k x^{k+r} = 0 \end{equation*}

Form the indicial equation

To form the indicial equation, look at the lowest power of x in the equation (\(x^{r-2}\)) and its coefficient: \begin{equation*} 2r(r-1)a_0 + ra_0 - a_0 = 0 \end{equation*} Factor out \(a_0\) and simplify the equation: \begin{equation} a_0(2r^2 - r - 1) = 0 \end{equation} For non-trivial solutions, \(a_0 \neq 0\). So, the indicial equation is: \begin{equation} 2r^2 - r - 1 = 0 \end{equation}

Solve the indicial equation

Solve the indicial equation for \(r\): \begin{equation} r = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \end{equation} So, the exponents are \(r_1 = 1\) and \(r_2 = -\frac{1}{2}\).

Conclusion

In conclusion, we found that the regular singular points of the given differential equation are \(x_1 = 0\) and \(x_2 = -2\). We applied the Frobenius method and found the indicial equation to be \(2r^2 - r - 1 = 0\). Solving this equation, we found the exponents at the singularity: \(r_1 = 1\) and \(r_2 = -\frac{1}{2}\).

Key concepts

Indicial Equation

In the context of solving differential equations with regular singular points, the indicial equation is crucial to understanding the behavior of solutions near the singularity. It's derived from applying the Frobenius method and looking at the lowest power of the independent variable—often denoted by x—in the series solution.
When we solve a differential equation using the Frobenius method, we assume a solution of the form \( y(x) = \sum_{k=0}^\infty a_k x^{k+r} \) where \( r \) is the exponent that we need to determine. Inserting the derivatives of this assumed solution into our differential equation and then equating the coefficients of the lowest power of \( x \) to zero yields the indicial equation. For the illustrative equation \( 2r^2 - r - 1 = 0 \), the roots \( r_1 = 1 \) and \( r_2 = -\frac{1}{2} \) indicate possible behaviors of the series solution around the singularity.

Importance of Indicial Equation

• It determines the possible exponents \( r \) in the Frobenius series.
• Provides insight into the nature of the solution near the singularity.
• Helps in constructing a power series solution for the differential equation.

Understanding how to find and solve the indicial equation is vital to unlocking the general solutions to complex differential equations with singular points.

Frobenius Method

The Frobenius method is a powerful technique used for finding solutions to linear differential equations around regular singular points. This method generalizes the power series solution by allowing the exponent in the series to be any real number, which is particularly useful when the standard power series are not applicable.
A key step in the Frobenius method is to write the series solution in a form that includes a summation and an undetermined exponent \( r \), as \( y(x) = \sum_{k=0}^\infty a_k x^{k+r} \) where \( a_k \) are the coefficients to be determined. Differentiating this series term-by-term, substituting into the differential equation, and equating coefficients of like powers of \( x \) allow us to determine the \( a_k \) coefficients and the exponent \( r \) through the indicial equation.
While applying this method to the provided exercise, we can derive the indicial equation and consequently determine the series' coefficients that would represent the solution around the singular points. Each series solution corresponds to a possible value of \( r \) derived from the indicial equation.

Role of the Frobenius Method

• It extends the power series method to include solutions about singular points.
• Allows the determination of coefficients in the series based on the differential equation's behavior near singularities.
• Facilitates understanding of the nature of solutions that otherwise would not be captured with standard methods.

Mastering the Frobenius method enables one to tackle a broader range of problems in differential equations where singular points play a significant role.

Differential Equations

Understanding differential equations is essential for describing various phenomena in physics, engineering, economics, and even biology. A differential equation is an equation that involves an unknown function and its derivatives, and it specifies how a quantity changes in relation to another. The equation we are considering, \( 2 x(x+2) y^{\prime \prime}+y^{\prime}-x y=0 \), is an example of a linear second-order differential equation.
Generally, the solutions to differential equations are functions that satisfy the given relation among the function and its derivatives. These solutions can be straightforward, such as in the case of separable or exact equations, or more complex, demanding methods like the Frobenius method when dealing with singular points.
Differential equations can be broadly classified into two categories:

• Ordinary Differential Equations (ODEs): Equations involving derivatives with respect to a single independent variable, as is the case in our exercise.
• Partial Differential Equations (PDEs): Equations involving partial derivatives with respect to multiple independent variables.

Finding the solutions often requires understanding the nature of the equation, identifying singular points, and employing appropriate methods, such as separation of variables, power series techniques, or advanced methods like the Frobenius method for more intricate problems.
Differential equations are indispensable for modeling real-world systems, making it crucial for students to develop proficiency in various methods for solving them, especially when dealing with equations showcasing singular points or particular behaviors.