Question

In Problems 15–24,x = 0is a regular singular point of the given dif-

differential equation. Show that the indicial roots of the singularity do not differ by an integer. Use the method of Frobenius to obtain two linearly independent series solutions about x=0. Form the general solution on$\left(0,\infty \right)$.

localid="1664862755026" ${\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\left(x^2-\frac{4}{9}\right)\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

Therefore, the solution is:

$y=C_1{x}^{2/3}\left[1-\frac{3}{20}{x}^2+\frac{9}{1280}{x}^4+\dots \right]+C_2{x}^{-2/3}\left[1-\frac{3}{4}{x}^2+\frac{9}{128}{x}^4+\dots \right]$

Step-by-step solution

Given Information.

The given problem is;

$x^{2}y\text{'}\text{'}+xy\text{'}+\left(x^2-\frac{4}{9}\right)y=0$

Use Differentiation.

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;

$x^{2}y\text{'}\text{'}+xy\text{'}+\left(x^2-\frac{4}{9}\right)y=0$

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

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

We find the first and second derivatives.

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

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

Substitute the equation.

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

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=x^2\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2}+x\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r-1}+\left(x^2-\frac{4}{9}\right)\sum _{n=0}^{\infty }{c}_n{x}^{n+r} \\ =\underset{a_0\to x^r}{\underbrace{\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r}}}+\underset{a_0\to x^r}{\underbrace{\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r}}}+\underset{a_0\to x^{r+2}}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+r+2}}}-\underset{a_0\to x^r}{\underbrace{\sum _{n=0}^{\infty }\frac{4}{9}{c}_n{x}^{n+r}}} \end{gathered}$$

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

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left[c_{0}r\right(r-1)x^r+c_1(r+1\left)\right(r)x^{r+1}+\underset{a_2\to x^{r+2}}{\underbrace{\left.\sum _{n=2}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r}\right]}}+\left[c_{0}r{x}^r+c_1(r+1)x^{r+1}\right. \\ +\underset{a_2\to x^{r+2}}{\underbrace{\left.\sum _{n=2}^{\infty }{c}_n(n+r)x^{n+r}\right]}}+\underset{a_0\to x^{r+2}}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+r+2}}}-[\frac{4}{9}{c}_0{x}^r+\frac{4}{9}{c}_1{x}^{r+1}+\underset{a_2\to x^{r+2}}{\underbrace{\left.\sum _{n=2}^{\infty }{c}_n{x}^{n+r}\right]}} \end{gathered}$$

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

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=c_{0}r(r-1)x^r+c_1(r+1)\left(r\right)x^{r+1}+\sum _{n=2}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r}+c_{0}r{x}^r+c_1(r+1)x^{r+1} \\ +\sum _{n=2}^{\infty }{c}_n(n+r)x^{n+r}+\sum _{n-2=0}^{\infty }{c}_{n-2}{x}^{(n-2)+r+2}-\frac{4}{9}{c}_0{x}^r-\frac{4}{9}{c}_1{x}^{r+1}-\sum _{n=2}^{\infty }{c}_n{x}^{n+r} \\ =c_{0}r(r-1)x^r+c_1(r+1)\left(r\right)x^{r+1}+\sum _{n=2}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r}+c_{0}r{x}^r+c_{10}(r+1)x^{r+1} \\ +\sum _{n=2}^{\infty }{c}_n(n+r)x^{n+r}+\sum _{n=2}^{\infty }{c}_{n-2}{x}^{n+r}-\frac{4}{9}{c}_0{x}^r-\frac{4}{9}{c}_1{x}^{r+1}-\sum _{n=2}^{\infty }{c}_n{x}^{n+r} \end{gathered}$$

The coefficient of terms:

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

$y=C_1{x}^{2/3}\left[1-\frac{3}{20}{x}^2+\frac{9}{1280}{x}^4+\dots \right]+C_2{x}^{-2/3}\left[1-\frac{3}{4}{x}^2+\frac{9}{128}{x}^4+\dots \right]$

This implies that;

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=\left[r(r-1)+r-\frac{4}{9}\right]c_0{x}^r+\left[(r+1)\left(r\right)+(r+1)-\frac{4}{9}\right]c_1{x}^{r+1}+\sum _{n=2}^{\infty }\left[c_n(n+r)(n+r-1)+c_n(n+r)+c_{n-2}-\frac{4}{9}{c}_n\right]x^{n+r} \\ =\left[r^2-r+r-\frac{4}{9}\right]c_0{x}^r+\left[r^2+r+r+1-\frac{4}{9}\right]c_1{x}^{r+1}+\sum _{n=2}^{\infty }\left[c_n(n+r)(n+r-\backslash \mathrm{not}1+\backslash \mathrm{not}1)+c_{n-2}-\frac{4}{9}{c}_n\right]x^{n+r} \\ =\left[r^2-\frac{4}{9}\right]c_0{x}^r+\left[r^2+2r+\frac{5}{9}\right]c_1{x}^{r+1}+\sum _{n=2}^{\infty }\left[c_n\left({(n+r)}^2-\frac{4}{9}\right)+c_{n-2}\right]x^{n+r} \\ =0 \end{gathered}$$

This implies that;

$r^2-\frac{4}{9}=0\Rightarrow \left(r-\frac{2}{3}\right)\left(r+\frac{2}{3}\right)=0$

Therefore,

$r_1=\frac{2}{3}$and $r_2=\frac{-2}{3}$

$c_n\left({(n+r)}^2-\frac{4}{9}\right)+c_{n-2}=0\Rightarrow c_n=\frac{-c_{n-2}}{{(n+r)}^2-\frac{4}{9}}$

The coefficient of the series.

For r₁=2/3 we have

$y=\sum _{n=0}^{\infty }{c}_n{x}^{n+2/3}=x^{2/3}\sum _{n=0}^{\infty }{c}_n{x}^n$

And the recurrence relation is;

$$\begin{gathered} c_n=\frac{-c_{n-2}}{{\left(n+\frac{2}{3}\right)}^2-\frac{4}{9}} \\ =\frac{-9c_{n-2}}{{(3n+2)}^2-4} \dots \left(5\right) \end{gathered}$$

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

At, n = 2

$$\begin{gathered} c_2=\frac{-9c_0}{8^2-4} \\ =\frac{-3c_0}{20} \end{gathered}$$

At , n = 3

n = 3, c₃= 0

$$\begin{gathered} c_4=\frac{-9c_2}{192} \\ =\frac{-3\left(-3c_0/20\right)}{64} \\ =\frac{9c_0}{1280} \end{gathered}$$

Expand the series

$$\begin{gathered} y_1=x^{2/3}\left[c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \right] \\ =x^{2/3}\left[c_0+0·x-\frac{3}{20}{x}^2+0·x^3+\frac{9}{1280}{c}_0{x}^4+\dots \right] \\ \text{=}{c}_0{x}^{2/3}\left[1-\frac{3}{20}{x}^2+\frac{9}{1280}{x}^4+\dots \right] \end{gathered}$$

Which is the series solution for r = 2/3

For r₂ = -2/3 we have;

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^{n-2/3} \\ =x^{-2/3}\sum _{n=0}^{\infty }{c}_n{x}^n \end{gathered}$$

And the recurrence relation is;

$$\begin{gathered} c_n=\frac{-c_{n-2}}{{\left(n-\frac{2}{3}\right)}^2-\frac{4}{9}} \\ \text{=}\frac{-9c_{n-2}}{{(3n-2)}^2-4} \dots \left(7\right) \end{gathered}$$

Now, we find the coefficients of the series, using (4) and (7).

At n = 2

$$\begin{gathered} c_2=\frac{-9c_0}{16-4} \\ =\frac{-9c_0}{12} \\ =\frac{-3c_0}{4} \end{gathered}$$

At n = 3, C₃= 0

For n = 4

$$\begin{gathered} c_4=\frac{-9c_2}{100-4} \\ =\frac{-9\left(-3c_0/4\right)}{96} \\ =\frac{9c_0}{128} \end{gathered}$$

Now, we expand the series as follows;

$$\begin{gathered} y_2=x^{-2/3}\left[c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \right] \\ =c_0{x}^{-2/3}\left[1-\frac{3}{4}{x}^2+\frac{9}{128}{x}^4+\dots \right] \dots \left(8\right) \end{gathered}$$

Which is the series solution for r₂=-2/3 After omitting we can obtain a general solution by using the superposition principle.

$y=C_1{x}^{2/3}\left[1-\frac{3}{20}{x}^2+\frac{9}{1280}{x}^4+\dots \right]+C_2{x}^{-2/3}\left[1-\frac{3}{4}{x}^2+\frac{9}{128}{x}^4+\dots \right]$