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 $\left(0,\infty \right)$

$\mathbf{27}\mathbf{.}{\mathbf{xy}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathbf{xy}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

$y=C_{1}x+C_2-1+xlnx+\frac{1}{2}{x}^2+\frac{1}{12}{x}^3+\dots$

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

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

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

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

We find the first and second derivative.

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

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

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

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

Find the coefficients of the terms

Our aim, now, is to make the summation index and the power of x 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

$f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=c_{0}r(r-1)x^{r-1}+\underset{a_1\to x^r}{\underbrace{\underset{n=1}{\underbrace{\infty }}{c}_n(n+r)(n+r-1)x^{n+r-1}}}\underset{a_0\to x^r}{\underbrace{c_{n=0}^{\infty }(n+r)x^{n+r}}}+\underset{a_0\to x^r}{\underbrace{c_n{x}^{n+r}}}$

Now, we do the shift for the third series from n to n-2 yields

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=c_{0}r(r-1)x^{r-1}+c_{n+1}\left(\right(n+1)+r)\left(\right(n+1)+r-1)x^{(n+1)+r-1}{c}_{n=0}^{\infty }(n+r)x^{n+r}{+}_{n=0}^{\infty }{c}_n{x}^{n+r} \\ \left.=r(r-1)c_0{x}^{r-1}+c_{n=0}{c}_{n+1}(n+r+1)(n+r)-c_{n}n+r-1\right)x^n \\ {}^{-1}+c_{n=0}^{\infty }{c}_{n+1}(n+r+1)(n+r)x^{n+r}-c_{n=0}^{\infty }{c}_n(n+r)x^{n+r}{+}_{n=0}^{\infty }{c}_n{x}^{n+r} \end{gathered}$$

$r(r-1)=r_1=1and{r}_2=0\left.c_{n+1}(n+r+1)(n+r)-c_{n}n+r-1\right)=0\Rightarrow c_{n+1}=$

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

For r₁ = 1 we have

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

and the recurrence relation is

$c_{n+1}=\frac{c_n(n+\backslash \mathrm{not}1-\backslash \mathrm{not}1)}{(n+1+1)(n+1)}=\frac{c_{n}n}{(n+2)(n+1)}$

Now, we find the coefficients of the series, using the recurrence relation.

$\text{at}n=0,c_1=\frac{c_0·0}{2}=\frac{0}{2}$

Final proof

Therefore, the solution for r₁= 1 becomes as follows

$$\begin{gathered} y_1==x\left[c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \right] \\ =x\left[c_0+0+0+\dots +0\right] \\ =c_{0}x\dots \left(4\right) \end{gathered}$$

For r₂ = 0

we have

$y=c_{n=0}^{\infty }{c}_n{x}^n$

and the recurrence relation is

$(n-11+\backslash \mathrm{not}1)(n-1)c_{n+1}=c_n(n-1-1)n(n-1)c_{n+1}=c_n(n-2)$

Therefore, we will not be able to obtain a second solution, using Forbinous Method, as

at $$\begin{gathered} n=0,c_0=0⋮c_n=0 \\ for\text{n}=0,1,2,\dots \end{gathered}$$

Recall (23) from the book, we can obtain a second solution as follows

$y_2\left(x\right)=y_1\left(x\right)\frac{e^{-}P\left(x\right)dx}{y_1^2\left(x\right)}dx\dots \left(5\right)$

We put the given differential equation in the standard form, yields

$-y^{\text{'}\text{'}}--y^{\text{'}}+\frac{1}{-y}=0$

$y^{\text{'}\text{'}}-y^{\text{'}}+\frac{1}{x}y=0$

Therefore, $P\left(x\right)=-1\dots \left(6\right)andQ\left(x\right)=\frac{1}{x}-+$

$y=C_{1}x+C_2-1+xlnx+\frac{1}{2}{x}^2+\frac{1}{12}{x}^3+\dots$