Question

In Problems 31 and 32, x = 0is a regular singular point of the given differential equation. Show that the indicial roots of the singularity differ by an integer. Use the recurrence relation found by the method of Frobenius first with the larger root r₁. How many solutions did you find? Next use the recurrence relation with the smaller root. How many solutions did you find?

31.$\mathbf{xy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{6}\mathbf{)}\mathbf{y}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

The answer is:$$\begin{gathered} y_1=x^7[1-\frac{1}{2}x+\frac{5}{36}{x}^2-\frac{1}{36}{x}^3+...] \\ {y}_2=[1-\frac{1}{2}x+\frac{1}{10}{x}^2-\frac{1}{120}{x}^3] \end{gathered}$$

Step-by-step solution

Definition of Frobenius method:

The Frobenius technique may be used to find a power series solution to a differential equation if p(z) and q(z) are analytic at 0 or if both of their limits at 0 are analytic elsewhere (and are finite).

Applying Frobenius method:

We state, without a proof, that if we have a differential equation, with the following standard form;

$y\text{'}\text{'}+P\left(x\right)y\text{'}+Q\left(x\right)=0$

which has a regular singular point at x =x₀, then we can use Forbenious Method to solve this differential equation. We have

$xy\text{'}\text{'}+\left(x--6\right)y\text{'}--3y=0$

which has a singular point at x= 0. We assume the solution, according to Forbenious, to be as follows:

$y=\sum _{n=0}^{\infty }{c}_n{x}^{n+r}...\left(1\right)$

We find the first and second derivative.

We find the first and second derivative.

$$\begin{gathered} y\text{'}=\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r-1}...\left(2\right) \\ y\text{'}\text{'}=\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2}...\left(3\right) \end{gathered}$$

$$\begin{gathered} f(y\text{'}\text{'},y\text{'},y)=x\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2}(x-6)\sum _{n=0}^{\infty }{c}_n(n+r)x^{n=r-1}-3\sum _{n=0}^{\infty }{c}_n{x}^{n+r} \\ =\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-1}+\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r}-\sum _{n=0}^{\infty }6c_n(n+r)x^{n+r-1}-\sum _{n=0}^{\infty }3c_n{x}^{n+r} \end{gathered}$$

Our aim, now, is to make the summation index and the power of x, for the four series, in the same phase. We, first, have to assure that the first terms,

for the four series, are raised to the same power, which is not the case. Therefore, we do the following

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=c_{0}r(r-1)x^{r-1}+\sum _{n-1=0}^{\infty }{c}_{n-1}\left(\right(n-1)+r)x^{(n-1)+r}-6c_{0}r{x}^{r-1}-\sum _{n=1}^{\infty }6c_n(n+r)x^{n+r-1}-\sum _{n-1=0}^{\infty }3c_{n-1}{x}^{(n-1)+r} \\ =c_{0}r(r-1)x^{r-1}+\sum _{n=1}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-1}+\sum _{n=1}^{\infty }{c}_{n-1}(n+r-1)x^{n+r-1}-6c_{0}r{x}^{r-1}-\sum _{n=1}^{\infty }6c_n(n+r)x^{n+r-1}-\sum _{n=1}^{\infty }3c_{n-1}{x}^{n+r-1} \end{gathered}$$

And the second solution, obtained from r₂ = 0 ,is

$$\begin{gathered} {\text{y}}_3{\text{=c}}_0\text{+}...{\text{+c}}_6{\text{x}}^6+c_7{x}^7+c_8{x}^8+c_9{x}^9+c_{10}{x}^{10}+... \\ =0+c_7{x}^7-\frac{1}{2}{c}_7{x}^8+\frac{5}{36}{c}_7{x}^9-\frac{1}{36}{c}_7{x}^{10}+... \\ =c_7{x}^7[1-\frac{1}{2}x+\frac{5}{36}{x}^2-\frac{1}{36}{x}^3+...] \end{gathered}$$

Where c₀ and c₇are arbitrary constants.

Therefore, the solutions, obtained by using Forbinous method,are

$$\begin{gathered} y_1=x^7[1-\frac{1}{2}x+\frac{5}{36}{x}^2-\frac{1}{36}{x}^3+...] \\ {y}_2=[1-\frac{1}{2}x+\frac{1}{10}{x}^2-\frac{1}{120}{x}^3] \end{gathered}$$