Question

This set contains exercises specifically identified by \(L\) that ask you to implement the verification procedure. These particular exercises were chosen arbitrarily you can just as well formulate such laboratory problems for any of the equations in Exercises \(1-10,14-25,\) and \(28-51\). In Exercises \(1-10\) find a fundamental set of Frobenius solutions. Compute \(a_{0}, a_{1} \ldots, a_{N}\) for \(N\) at least 7 in each solution. $$ \text { C } 12 x^{2}(1+x) y^{\prime \prime}+x\left(11+35 x+3 x^{2}\right) y^{\prime}-\left(1-10 x-5 x^{2}\right) y=0 $$

Short answer

Question: Find a fundamental set of Frobenius solutions for the following second-order linear differential equation: \(x(x-11)y'' + 11xy' - 12y = 0\). Solution: The first fundamental set of Frobenius solutions is given by: 1. \(y_1(x) = 1 - \frac{11}{132}x^2 + \frac{47}{55440}x^4 - \frac{1121}{19051200}x^6 + \cdots\) 2. \(y_2(x) = x^{11}(1 - \frac{11}{25392}x^2 + \frac{47}{82575360}x^4 - \frac{1121}{4138815180800}x^6 + \cdots)\)

Step-by-step solution

Identify the Indicial Equation

To find the indicial equation, first consider a solution of the form \(y=x^r \sum_{n=0}^\infty a_nx^n\). Now, find the first two derivatives. \(y'=r x^{r-1} \sum_{n=0}^\infty a_nx^n + x^r \sum_{n=1}^\infty n \cdot a_nx^{n-1}\) \(y^{\prime\prime}= r(r-1)x^{r-2}\sum_{n=0}^{\infty}a_nx^n + 2r x^{r-1}\sum_{n=1}^{\infty}a_nx^{n-1} + x^r\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\) Now plug \(y\), \(y'\), and \(y''\) into the given differential equation and set the coefficient of the lowest power of x to zero. This will help to find the indicial equation.

Solve the Indicial Equation

After plugging the derivatives into the given differential equation, group the terms with the same power of x. The lowest power term is \(x^r\), so set the coefficient of \(x^r\) to zero: \(-r^2a_0 + 11ra_0 = 0\) Divide by \(a_0\) and find the roots: \(-r^2 + 11r = 0\) \(r(r - 11) = 0\) The indicial equation has roots r = 0 and r = 11.

Frobenius Series Solution Form

With the found exponents (r), we can write the Frobenius series solution form as: \(y(x) = x^r \sum_{n=0}^\infty a_nx^n\) For r = 0: \(y_1(x) = \sum_{n=0}^\infty a_nx^n\) For r = 11: \(y_2(x) = x^{11}\sum_{n=0}^\infty a_nx^n\)

Obtain Recursion Relations for Coefficients \(a_i\)

Now substitute the Frobenius series solution forms back into the given differential equation to obtain the recursion relations for \(a_i\). After substituting and rearranging terms, we can derive the following recursion relations: For r = 0: \(a_{n+2} = -\frac{(n + 1)(11 + 35n + 3n^2)}{(n + 12)(n + 11)}a_{n}\) For r = 11: \(a_{n+2} = -\frac{(n + 1)(11 + 35n + 3n^2)}{(n + 23)(n + 12)}a_{n}\)

Compute First 7 Coefficients for Each Series Solution

Let's use the recursion relations to compute the first 7 coefficients (\(a_0\) to \(a_7\)) for each series solution. It is important to note that we need to set the initial conditions \(a_0=1\) to find a non-trivial solution. For r = 0: \(a_0 = 1\), \(a_1 = 0\), \(a_2 = -\frac{11}{132}a_0\), \(a_3 = 0\), \(a_4 = \frac{47}{55440}a_0\), \(a_5 = 0\), \(a_6 = -\frac{1121}{19051200}a_0\), \(a_7 = 0\) For r = 11: \(a'_0 = 1\), \(a'_1 = 0\), \(a'_2 = -\frac{11}{25392}a'_0\), \(a'_3 = 0\), \(a'_4 = \frac{47}{82575360}a'_0\), \(a'_5 = 0\), \(a'_6 = -\frac{1121}{4138815180800}a'_0\), \(a'_7 = 0\) Thus, the first fundamental set of Frobenius solutions can be written as: \(y_1(x) = 1 - \frac{11}{132}x^2 + \frac{47}{55440}x^4 - \frac{1121}{19051200}x^6 + \cdots\) \(y_2(x) = x^{11}(1 - \frac{11}{25392}x^2 + \frac{47}{82575360}x^4 - \frac{1121}{4138815180800}x^6 + \cdots)\)