Question

Show that the given differential equation has a regular singular point at \(x=0 .\) Determine the indicial equation, the recurrence relation, and the roots of the indicial equation. Find the series solution \((x>0)\) corresponding to the larger root. If the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also. \(x^{2} y^{\prime \prime}+x y^{\prime}+(x-2) y=0\)

Short answer

The regular singular point of the given differential equation is x = 0. 2. What is the indicial equation for this problem? The indicial equation is r^2 a_0 = 0. 3. What are the roots of the indicial equation? The roots of the indicial equation are r = 0 (with only one root in this case). 4. What is the recurrence relation for the problem? The recurrence relation for the problem is a_{n+1} = (2a_{n-1}) / (n(n+2)). 5. What is the series solution corresponding to the larger root? The series solution corresponding to the larger root (r=0) is y(x) = a_0(1 + (2x)/(1⋅3) + (2x^2)/(2⋅4) + (2x^3)/(3⋅5) + …), where a_0 is an arbitrary non-zero constant.

Step-by-step solution

Frobenius Method

First, let's assume a solution for the given differential equation of the form \(y(x) = x^r \sum_{n=0}^\infty a_n x^n\). Then, we need to compute \(y'(x)\) and \(y''(x)\). The first derivatives are: $y'(x) = x^r \sum_{n=0}^\infty n\cdot a_n x^{n-1} = \sum_{n=0}^\infty n\cdot a_n x^{n+r-1}$ The second derivatives are: \(y''(x) = \sum_{n=0}^\infty (n-1)\cdot n\cdot a_n x^{n+r-2}\)

Substituting and simplifying

Now, we will substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the given differential equation: $x^2 y'' + xy' + (x-2)y = x^{r-2}\sum_{n=1}^\infty n(n-1) a_n x^{n} + x^{r-1}\sum_{n=1}^\infty n a_n x^{n} +(x-2)x^{r}\sum_{n=0}^\infty a_n x^{n} =0$ Then, we will equate the coefficients of like powers of \(x\).

Formulating the indicial equation

To find the indicial equation, let's examine the coefficient of \(x^{r-1}\): \(r(r-1)a_0 + r a_0 = 0 \qquad\) (Indicial equation) This simplifies to: \(r^2 a_0 = 0\) Since \(a_0 \neq 0\), we know \(r = 0\).

Recurrence relation

Now, we will find the recurrence relation for \(n \geq 1\). The original equation coefficients are: \(x^{r-2} [(n+1)^2 a_{n+1} + n^2 a_n] + x^{r-1} [(n+1) a_{n+1} + na_n] + x^r [a_n - 2a_{n-1}]=0\) We will equate the coefficients of like powers of \(x\). For \(n = 1\): \(2a_2+a_1 - 2a_0 = 0\) For \(n \geq 2\): \((n+1)^2 a_{n+1} + n^2 a_n +(n+1)a_{n+1} + na_n = 2a_{n-1}\) This becomes the recurrence relation: \(a_{n+1} = \frac{2a_{n-1}}{n(n+2)}\).

Finding the roots

We already found one root of the indicial equation: \(r=0\). Since the indicial equation is quadratic, it can have at most two roots. In this case, the roots are equal, and they do not differ by an integer.

Series solution

Now, we will find the series solution \((x>0)\) corresponding to the larger root \(r=0\). We obtain the coefficients using the recurrence relation: \(a_{1} = \frac{2a_{0}}{1\cdot 3}\) \(a_{2} = \frac{2a_{1}}{2\cdot 4}\) \(a_{3} = \frac{2a_{2}}{3\cdot 5}\) \(\vdots\) Now, combining the coefficients with the assumed solution form, we get: \(y(x) = a_0\left(1 + \frac{2x}{1\cdot 3} + \frac{2x^2}{2\cdot 4} + \frac{2x^3}{3\cdot 5} + \cdots\right)\) Here, \(a_0\) can be any arbitrary non-zero constant. This is the series solution \((x>0)\) corresponding to the larger root. Since the roots are equal, we do not have to find the series solution corresponding to the smaller root.

Key concepts

Indicial Equation

When solving differential equations with a regular singular point, one of the pivotal steps is to determine the indicial equation. This equation helps us find the initial exponents of the power series solution, which are crucial for identifying the solution's behavior near the singular point. The indicial equation is derived from substituting a trial solution of the form \(y(x) = x^r \sum_{n=0}^\infty a_n x^n\) into the differential equation and focusing on the lowest power of \(x\). By examining the coefficient of this lowest power, we formulate the indicial equation. In our specific problem, this process leads to the simplified equation: \[r(r-1)a_0 + ra_0 = 0,\] which further simplifies to \(r^2 a_0 = 0\). Since \(a_0\) is non-zero to ensure a non-trivial solution, we conclude that \(r = 0\).Thus, the indicial equation not only reveals the possible values of \(r\) but also provides essential initial conditions for further solving the differential equation using the Frobenius method.

Recurrence Relation

The recurrence relation arises from the process of matching coefficients in the power series expansion of the differential equation. Once we've determined the indicial equation and have values for \(r\), the next step is to find how the coefficients \(a_n\) relate to each other.In our problem, the original differential equation is substituted with the series expression. By aligning coefficients for each power of \(x\), we derive relations among the coefficients. For instance:

• At \(n = 1\), we find the equation \(2a_2 + a_1 - 2a_0 = 0\).
• For \(n \geq 2\), the equation goes \((n+1)^2 a_{n+1} + n^2 a_n + (n+1)a_{n+1} + na_n = 2a_{n-1}\).

This simplifies to the recurrence relation:\[a_{n+1} = \frac{2a_{n-1}}{n(n+2)}.\]Recurrence relations like these allow us to calculate any coefficient \(a_n\) based on the earlier term \(a_{n-1}\). Understanding them is crucial for building the full series solution.

Regular Singular Point

A regular singular point is a specific type of irregularity in a differential equation where certain methods, like the Frobenius method, can be applied effectively. A singular point \(x = x_0\) is termed 'regular' for a linear differential equation of the form \[x^2y'' + xpy' + qy = 0\]if both \(x^2p(x)\) and \(x^2q(x)\) remain analytic at \(x_0\). This ensures that, although the point \(x_0\) might introduce complications, the solutions around it maintain enough regularity to be expressed as a power series multiplied by \(x^r\).In the present differential equation, substituting \(x = 0\) reveals it as a regular singular point. Consequently, we can reassuringly proceed with the Frobenius method, allowing us to explore power series solutions and identify behavior around this pivotal location.