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.