Question

x = 0 is a regular singular point of the given differential equation. Show that the indicial roots of the singularity differ by an integer. Use the method of Frobenius to obtain at least one series solution about x = 0. Use (23) where necessary and a CAS, if instructed, to find a second solution. Form the general solution on $\left(0,\infty \right)$

localid="1664863875130" $\mathbf{25}\mathbf{.}{\mathbf{xy}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{xy}\mathbf{=}\mathbf{0}$

Short answer

$y=C_1^{\infty }\frac{x^{2n}}{(2n+1)!}+C_2{x^{-1}}_{n=0}^{\infty }\frac{x^{2n}}{\left(2n\right)!}$

Step-by-step solution

To Find the differential equation and use forbenious 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

$x{y}^{\text{'}\text{'}}+2y^{\text{'}}-xy=0$

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

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

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

We substitute (1), (2) and (3) into the given differential equation, yields

$f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=x{c}_n(n+r)(n+r-1)x^{n+r-2}+2c_n(n+r)x^{n+r-1}-x{c}_{n=0}{x}^{n+r}$

$=\underset{a_0\to x^{r}1}{\underbrace{n_{n=0}^{\infty }}}{c}_n(n+r)(n+r-1)x^{n+r-1}+\underset{a_0\to x^{r-1}}{\underbrace{\infty }}+\underset{a_0\to x^{r+1}}{\underbrace{n=0}}$

Find the coefficients of the terms

Our aim, now, is to make the summation index and the power of for the three series, in the same phase. We, first, have to assure that the first terms, for the three series, are raised to the same power, which is not the case. Therefore, we do the following

$f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=c_{0}r(r-1)x^{r-1}+c_1(r+1)\left(r\right)x^r+\underset{a_2\to x^{r+1}}{\underbrace{\underset{n=2}{\underbrace{\infty }}{c}_n(n+r)(n+r-1)x^{n+r-1}}}+2c_{0}r{x}^{r-1}+2c_1(r+1)x^r$

$+\underset{a_2\to x^{r+1}}{\underbrace{n=2c_n(n+r)x^{n+r-1}}}-\underset{a_0\to x^{r+1}}{\underbrace{n=0}}$

Now, we do the shift for the third series from yields

$f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=c_{0}r(r-1)x^{r-1}+c_1(r+1)\left(r\right)x^r+c_n(n+r)(n+r-1)x^{n+r-1}++2c_{0}r{x}^{r-1}+2c_1(r+1)x^r$

${+}_{n=2}^{\infty }2c_n(n+r)x^{n+r-1}{-}_{n-2=0}^{\infty }{c}_{n-2}{x}^{(n-2)+r+1}$

$=c_{0}r(r-1)x^{r-1}+c_1(r+1)\left(r\right)x^r{+}_{n=2}{c}_n(n+r)(n+r-1)x^{n+r-1}+2c_{0}r{x}^{r-1}+2c_1(r+1)x^r{+}_{n=2}^{\infty }2c_n(n+r)x^{n+r-1}-c_{n=2}^{\infty }{c}_{n-2}{x}^{n+r-1}$

Now, we sum the coefficients of the terms, which are raised to the same power together, yields

$f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=r(r-1)+2r{c}_0{x}^{r-1}+(r+1)\left(r\right)+2(r+1)c_1{x}^r+c_n(n+r)(n+r-1)+2c_n(n+r)-c_{n-2}{x}^{n+r-1}$

$=r^2-r+2r{c}_0{x}^{r-1}+r^2+r+2r+2c_1{x}^r+{\infty }_{n=2}^{\infty }{c}_n(n+r)(n+r-1+2)-c_{n-2}{x}^{n+r-1}$

$r^2+r{c}_0{x}^{r-1}+r^2+3r+2c_1{x}^r+c_{n=2}{c}_n(n+r)(n+r+1)-c_{n-2}{x}^{n+r-1}=0$

$$\begin{gathered} \text{whichimpliesthat}{r}^2+r=0r(r+1)=0 \\ {r}_1=0and{r}_2=-1 \\ {c}_1=0\dots \left(4\right) \end{gathered}$$

$c_n(n+r)(n+r+1)-c_{n-2}=0\Rightarrow c_n=\frac{c_{n-2}}{(n+r)(n+r+1)}$

For r₁= 0 we have

$y{=}_{n=0}^{\infty }{c}_n{x}^n$and the recurrence relation is

$c_n=\frac{c_{n-2}}{n(n+1)}\dots$

Final proof

Now, we find the coefficients of the series, using (4) and (5).

$$\begin{gathered} atn=2,c_2=\frac{c_0}{6} \\ =\frac{c_0}{3!} \end{gathered}$$

$\&n=3,c_3=0n=4,c_4=\frac{c_2}{20}=\frac{\left(c_0/6\right)}{20}=\frac{c_0}{120}=\frac{c_0}{5!}{c}_{2n}=\frac{c_0}{(2n+1)!}$

We expand the series solution.

$$\begin{gathered} y_1=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \\ =c_0+0\times x+\frac{c_0}{3!}{x}^2+0\times x^3+\frac{c_0}{5!}{x}^4+\dots \\ =c_0+\frac{c_0}{3!}{x}^2+\frac{c_0}{5!}{x}^4+\dots \\ =\frac{c_0}{(2n+1)!}{x}^{2n} \\ ={}_{n=0}{}^{\backslash \mathrm{currency}}\frac{x^{2n}}{(2n+1)!} \end{gathered}$$

which is the series solution for r₁ = 0 . Note that, c₁ = c₃ = c₅ = ... 0

For 2 = -1

For r₂= - 1 we have

$y{=}_{n=0}^{\infty }{c}_n{x}^{n-1}={x^{-1}}_{n=0}^{\infty }{c}_n{x}^n$

and the recurrence relation is

$=\frac{c_{n-2}}{(n-1)(n-1+1)}=\frac{c_{n-2}}{(n-1)n}\dots \left(7\right)$

$$\begin{gathered} Now,wefindthecoefficientsoftheseries,u\sin g\left(4\right)and\left(7\right). \\ atn=2, \\ {c}_2=\frac{c_0}{2!}nn=3, \\ {c}_3=0n=4, \end{gathered}$$

$$\begin{gathered} c_4=\frac{c_2}{4\times 3}n{c}_4=\frac{c_2}{4\times 3} \\ {c}_5=0n=6, \\ {c}_6=\frac{c_4}{6\times 5}=\frac{c_0}{6!} \end{gathered}$$

And the series solution, for $x=0$, is

role="math" ${\mathrm{y}}_2={{\mathrm{x}}^{-1}}_{\mathrm{n}=0}^{\infty }\frac{{\mathrm{c}}_0}{\left(2\mathrm{n}\right)!}{\mathrm{x}}^{2\mathrm{n}}={\mathrm{c}}_0{{\mathrm{x}}^{-1}}_{\mathrm{n}=0}^{\infty }\frac{{\mathrm{x}}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}$$c_{2n}=\frac{c_0}{\left(2n\right)!}$