Problem 4 - Differential Equations and Linear Algebra

Question

Determine the roots of the indicial equation of the given differential equation. Also obtain the general form of two linearly independent solutions to the differential equation on an interval \((0, R) .\) Finally, if \(r_{1}-r_{2}\) equals a positive integer, obtain the recurrence relation and determine whether the constant \(A\) in $$y_{2}(x)=A y_{1}(x) \ln x+x^{r_{2}} \sum_{n=0}^{\infty} b_{n} x^{n}$$ is zero or nonzero. $$x^{2} y^{\prime \prime}+x^{2} y^{\prime}-(2+x) y=0$$

Step-by-Step Solution

Answer

The roots of the indicial equation are \(r_1 = 0\) and \(r_2 = -1\). The general form of two linearly independent solutions is given by \(y_1(x) = x^{r_1} \sum_{n=0}^{\infty} a_n x^n\) and \(y_2(x) = A y_1(x) \ln(x) + x^{r_2} \sum_{n=0}^{\infty} b_n x^n\). The recurrence relation for the coefficients \(a_n\) and \(b_n\) is \((r+n+1)(r+n)(a_{n+1}-b_{n+1}) + (r+n-2)(a_n-b_n) = 0\). Since the given differential equation is inhomogeneous, the constant \(A\) must be nonzero.

Step 1: Convert the given equation to the normalized form

We first need to convert the given equation into its normalized form by dividing through by \(x^2\): \[ y^{\prime\prime} + y^{\prime} - \frac{2+x}{x^2} y = 0. \]

Step 2: Identify the indicial equation

Now let's assume a solution in the power series form: \[ y(x) = \sum_{n=0}^{\infty} a_n x^{r+n}, \] with the derivatives \[ y'(x) = \sum_{n=0}^{\infty} (r+n) a_n x^{r+n-1}, \] and \[ y''(x) = \sum_{n=0}^{\infty} (r+n)(r+n-1) a_n x^{r+n-2}. \]

Step 3: Substitute into the normalized equation and simplify

Substitute the power series and its derivatives into the normalized equation: \[ \sum_{n=0}^{\infty} (r+n)(r+n-1) a_n x^{r+n-2} + \sum_{n=0}^{\infty} (r+n) a_n x^{r+n-1} - \sum_{n=0}^{\infty} \left( \frac{2}{x^2} + \frac{1}{x} \right) a_n x^{r+n} = 0. \] To combine the series, we need to equalize the powers of x. We do this by shifting indices as follows: \[ \sum_{n=-1}^{\infty} (r+n+1)(r+n) a_{n+1} x^{r+n} + \sum_{n=0}^{\infty} (r+n) a_n x^{r+n} - \sum_{n=0}^{\infty} \left( 2a_n x^{r+n-2} + a_n x^{r+n-1} \right) = 0. \] Now we can combine the sums. Also, we will separate the n=0 term in the first sum since this requires the indicial equation: \[ \left[ (r+1)(r) a_1 x^{r-1} + 2a_0 x^{r-2} + a_0 x^{r-1} \right] + \sum_{n=1}^{\infty} x^{r+n} \left[ (r+n+1)(r+n) a_{n+1} + (r+n) a_n - 2a_n - a_n \right] = 0. \]

Step 4: Solve the indicial equation

The indicial equation is obtained by setting the coefficient of the term \(x^{r-1}\) equal to zero: \[ (r+1)(r) a_1 + a_0 = 0. \] We are assuming \(a_1 \ne 0\), so: \[ r^2 + r = 0, \] which has roots \[ r_1 = 0, \quad r_2 = -1. \]

Step 5: Obtain the general form of two linearly independent solutions and the recurrence relation

Since \(r_1 - r_2 = 1\), a positive integer, we need to consider the following general form of two linearly independent solutions: \[ y_1(x) = x^{r_1} \sum_{n=0}^{\infty} a_n x^n, \qquad y_2(x) = A y_1(x) \ln(x) + x^{r_2} \sum_{n=0}^{\infty} b_n x^n. \] Substituting these back into the combined series equation, we obtain the recurrence relation for the coefficients \(a_n\) and \(b_n\): \[ (r+n+1)(r+n)(a_{n+1}-b_{n+1}) + (r+n-2)(a_n-b_n) = 0. \]

Step 6: Determine whether the constant \(A\) is zero or nonzero

Since the logarithmic term in \(y_2(x)\) is multiplied by \(A\), we need to determine if \(A\) is zero or nonzero when \(r_1-r_2\) is a positive integer. If \(A=0\), then \(y_2(x)\) becomes a purely polynomial solution. However, we observe that since the differential equation involves an inhomogeneous term \(-\frac{2+x}{x^2} y = 0\), the logarithmic term is essential for the solution matching the second-order linear differential equation. Therefore, \(A\) must be nonzero.

Key Concepts

Indicial Equation

The indicial equation is a fundamental part of solving a differential equation using the Frobenius method. It helps in determining the leading behavior of power series solutions near a singular point. To understand it, let's first consider a differential equation in the form:

• Find an equation involving only the coefficient of the lowest power term from a series expansion.
• For this equation, solve for the powers that make the coefficient zero.

In the given problem, the differential equation is transformed, and a series expansion is assumed. The indicial equation emerges from setting the coefficient of the lowest power of x to zero, leading to roots known as indicial roots. For this specific problem, the indicial equation is:\[ r^2 + r = 0 \]which reveals its roots \( r_1 = 0 \) and \( r_2 = -1 \). These roots are critical in developing the overall solution structure.

General Solution

After finding the indicial equation, our next goal is to find the general solution for the differential equation. This solution will help us understand how the differential equation behaves for different functions.

• Identify two linearly independent solutions around the point of interest, often the singular point.
• Combine them to create a general solution.

In this context, the roots \( r_1 = 0 \) and \( r_2 = -1 \) from the indicial equation indicate that two linearly independent solutions can be constructed as follows:\[ y_1(x) = x^{r_1} \sum_{n=0}^{\infty} a_n x^n \]\[ y_2(x) = A y_1(x) \ln(x) + x^{r_2} \sum_{n=0}^{\infty} b_n x^n \]The general solution combines these solutions and considers any connection between them, establishing that the solution covers a wide range of potential behaviors.

Linear Independence

Linear independence is a key concept in differential equations, essential for constructing a general solution. It ensures that no solution can be expressed as a linear combination of other solutions. This is critical for guaranteeing that a complete solution space is spanned.

• Two solutions \(y_1(x)\) and \(y_2(x)\) are linearly independent if their Wronskian is not zero.
• This ensures the methods capture all nuances of the solution space.

In our exercise, the solutions \( y_1(x) \) and \( y_2(x) \) are tailored such that they provide a comprehensive exploration of solution possibilities. This is secured by the fact that they incorporate different exponentials and logarithmic components linked to the roots of the indicial equation. Thus, the solutions can't be simplified into each other, confirming their independence.

Recurrence Relation

Recurrence relations play a vital role when working with series solutions, especially in calculating coefficients of terms systematically. They offer a step-by-step approach to finding coefficients necessary for constructing the series representation of solutions.

• Recurrence relations derive from substituting series into differential equations and equating coefficients.
• They form an iterative process that determines each term based on previous ones.

For our differential equation scenario, the recurrence relation is established for coefficients \( a_n \) and \( b_n \), reflecting the relationships required for solving series expansions:\[ (r+n+1)(r+n)(a_{n+1}-b_{n+1}) + (r+n-2)(a_n-b_n) = 0 \]Through this recurrence relation, we systematically find the coefficients that define the power series, aiding in the solution construction for complex differential equations.