Question

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 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(0,$\mathbf{\infty }$)

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{x}}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

$y=C_1\left[1+\frac{1}{4}{x}^2+\right.\left.\frac{1}{48}{x}^4+\dots \right]+C_2\left[1+\frac{1}{4}{x}^2+\frac{1}{48}{x}^4+\dots \right]\left[-\frac{1}{2x}-\frac{1}{2}\text{ln}x+\frac{7}{96}{x}^2+\dots \right]$

Step-by-step solution

The differential equation

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

$y^{\text{'}\text{'}}+\frac{3}{x}{y}^{\text{'}}-2y=0$

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

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

We find the first and second derivative.

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

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

localid="1664864623160" $\begin{array}{rcl}\left({\mathrm{y}}^{\text{'}\text{'}},{\mathrm{y}}^{\text{'}},\mathrm{y}\right)& =& \sum _{\mathrm{n}=0}^{\infty }{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}+\frac{3}{\mathrm{x}}\sum _{\mathrm{n}=0}^{\infty }{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r}){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-1}-2\sum _{{\mathrm{a}}_{\mathrm{n}}}^{\infty }{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+\mathrm{r}}\\ & =& \underset{{\mathrm{a}}_0\to {\mathrm{x}}^{\mathrm{r}-2}}{\underbrace{\sum _{\mathrm{n}=0}^{\infty }{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}}}+\underset{{\mathrm{a}}_0\to {\mathrm{x}}^{\mathrm{r}}}{\underbrace{\sum _{\mathrm{n}=0}^{\infty }3{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r}){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}}}-\underset{}{\underbrace{\sum _{\mathrm{n}=0}^{\infty }2{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+\mathrm{r}}}}\end{array}$

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

localid="1664864639997" $$\begin{gathered} \left({\mathrm{y}}^{\text{'}\text{'}},{\mathrm{y}}^{\text{'}},\mathrm{y}\right)=\left[{\mathrm{c}}_0\mathrm{r}\right(\mathrm{r}-1){\mathrm{x}}^{\mathrm{r}-2}+{\mathrm{c}}_1(\mathrm{r}+1){\mathrm{rx}}^{\mathrm{r}-1}+\underset{{\mathrm{a}}_2\to {\mathrm{x}}^{\mathrm{r}}}{\underbrace{\left.\sum _{\mathrm{n}=2}^{\infty }{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}\right]}} \\ [{\mathrm{c}}_0\mathrm{r}(\mathrm{r}-1){\mathrm{x}}^{\mathrm{r}-2}+{\mathrm{c}}_1(\mathrm{r}+1){\mathrm{rx}}^{\mathrm{r}-1}+\underset{{\mathrm{a}}_2\to {\mathrm{x}}^{\mathrm{r}}}{\underbrace{\left.\sum _{\mathrm{n}=2}^{\infty }{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}\right]}} \end{gathered}$$

The third series

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

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=c_{0}r(r-1)x^{r-2}+c_1(r+1)r{x}^{r-1}+\sum _{n=2}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2} \\ +3c_{0}r{x}^{r-2}+3c_1(r+1)x^{r-1}+\sum _{n=2}^{\infty }3c_n(n+r)x^{n+r-2}-\sum _{n-2=0}^{\infty }2c_{n-2}{x}^{(n-2)+r} \\ =c_{0}r(r-1)x^{r-2}+c_1(r+1)r{x}^{r-1}+\sum _{n=2}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2} \\ +3c_{0}r{x}^{r-2}+3c_1(r+1)x^{r-1}+\sum _{n=2}^{\infty }3c_n(n+r)x^{n+r-2}-\sum _{n=2}^{\infty }2c_{n-2}{x}^{n+r-2} \end{gathered}$$

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

localid="1664864659714" $\begin{array}{rcl}\mathrm{f}\left({\mathrm{y}}^{\text{'}\text{'}},{\mathrm{y}}^{\text{'}},\mathrm{y}\right)& =& \left[\mathrm{r}\right(\mathrm{r}-1)+3\mathrm{r}]{\mathrm{c}}_0{\mathrm{x}}^{\mathrm{r}-2}+\left[\right(\mathrm{r}+1)\mathrm{r}+3(\mathrm{r}+1\left)\right]{\mathrm{c}}_1{\mathrm{x}}^{\mathrm{r}-1}+\sum _{\mathrm{n}=2}^{\infty }\left[{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1)+3{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})-2{\mathrm{c}}_{\mathrm{n}-2}\right]{\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}\\ & & \left[\mathrm{r}\right(\mathrm{r}-1)+3\mathrm{r}]{\mathrm{c}}_0{\mathrm{x}}^{\mathrm{r}-2}+\left[\right(\mathrm{r}+1)\mathrm{r}+3(\mathrm{r}+1\left)\right]{\mathrm{c}}_1{\mathrm{x}}^{\mathrm{r}-1}+\sum _{\mathrm{n}=2}^{\infty }\left[{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1)+3{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})-2{\mathrm{c}}_{\mathrm{n}-2}\right]{\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}\\ & =& \left[\mathrm{r}\right(\mathrm{r}+2\left)\right]{\mathrm{c}}_0{\mathrm{x}}^{\mathrm{r}-2}+\left[{\mathrm{r}}^2+4\mathrm{r}+3\right]{\mathrm{c}}_1{\mathrm{x}}^{\mathrm{r}-1}+\sum _{\mathrm{n}=2}^{\infty }\left[{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}+2)-2{\mathrm{c}}_{\mathrm{n}-2}\right]{\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}\\ & & \end{array}$

r(r+2) = 0

r₁= 0 and r₂= -2

c₁ = 0

localid="1664864686214" $\left.{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}+2)-2{\mathrm{c}}_{\mathrm{n}-2}=0\Rightarrow {\mathrm{c}}_{\mathrm{n}}=\frac{2{\mathrm{c}}_{\mathrm{n}-2}}{(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}+2)}[\text{wheren = 2},\text{3},\text{4},\dots \right]$

the two roots are distinct and differ by an integer

Therefore, the two roots are distinct and differ by an integer. We are left with calculating the series coefficients, using the recurrence relation for each case. Note that, you can't set c₀ and c₁ to be zero as the coefficients of the series are involved with c₀and c₁. You have to options;

For r₁ = 0, we have

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

and the recurrence relation is

$\left.c_n=\frac{2c_{n-2}}{n(n+2)}[\text{wheren = 2},\text{3},\text{4},\dots \right]$

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

$$\begin{gathered} n=2,c_2=\frac{2c_0}{8} \\ =\frac{c_0}{4} \\ n=3,c_3=0[\text{Recall}\left.{\text{c}}_{\text{-}}\text{1 = 0}\right] \\ n=4,c_4=\frac{2c_0}{24} \\ =\frac{\left(c_0/4\right)}{12} \\ =\frac{c_0}{48} \end{gathered}$$

Therefore, the solution for becomes as follows

$$\begin{gathered} y_1=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \\ =c_0+0·x+\frac{1}{4}{c}_0{x}^2+0·x^3+\frac{1}{48}{c}_0{x}^4+\dots \\ =c_0\left[1+\frac{1}{4}{x}^2+\frac{1}{48}{x}^4+\dots \right]\dots \left(4\right) \end{gathered}$$

For r₂ = -2, we have

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

and the recurrence relation is

$$\begin{gathered} c_n(n-2)n=2c_{n-2}\text{n}=2,3,4,\dots \\ {c}_0=0 \\ {c}_n=0 \end{gathered}$$

Therefore, we will not be able to obtain a second solution, using Forbinous Method. Recall (23) from the book, we can obtain a second solution as follows

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

For the given differential equation, we have

$$\begin{gathered} y^{\text{'}\text{'}}+\frac{3}{x}{y}^{\text{'}}-2y=0 \\ P\left(x\right)=\frac{3}{x}\dots \left(6\right)\text{and}Q\left(x\right)=-2 \end{gathered}$$

We substitute (6) and (4) into (5), after omitting c_{0}.

$$\begin{gathered} y_2\left(x\right)=y_1\left(x\right)\int \frac{e^{-\int (3/x)dx}}{y_1^2\left(x\right)}dx \\ =y_1\left(x\right)\int \frac{e^{-3\ln x}}{y_1^2\left(x\right)}dx \\ =y_1\left(x\right)\int \frac{e^{\ln x^{-3}}}{y_1^2\left(x\right)}dx\dots \left(7\right) \end{gathered}$$

Let

localid="1664864716073" $\begin{array}{rcl}\mathrm{u}& =& {\mathrm{e}}^{{\mathrm{lnx}}^{-3}}\Rightarrow \mathrm{lnu}={\mathrm{lne}}^{{\mathrm{lnx}}^{-3}}\\ & =& {\mathrm{lnx}}^{-3}\mathrm{lne}{\mathrm{lnx}}^{-3}\left|\text{Recall}{\mathrm{lnx}}^{\mathrm{y}}=\mathrm{ylnx}\right|\\ & =& [\text{Recall}\mathrm{lne}=1]\\ & & \end{array}$

Final proof

Therefore,

$$\begin{gathered} y_2\left(x\right)=y_1\left(x\right)\int \frac{y_1\left(x\right)\int \frac{x^{-3}}{y_1^2\left(x\right)}dx}{x^3{\left[1+\frac{1}{4}{x}^2+\frac{1}{48}{x}^4+\dots \right]}^2}dx \\ =y_1\left(x\right)\int \frac{1}{x^3\left[1+\frac{1}{2}{x}^2+\frac{5}{48}{x}^4+\dots \right]}dx\left[\text{UsingMultinomialTheoreminCombinatorics}\right] \\ =y_1\left(x\right)\int \frac{1}{\left[x^3+\frac{1}{2}{x}^5+\frac{5}{48}{x}^7+\dots \right]}dx \\ =y_1\left(x\right)\int \left[\frac{1}{x^3}-\frac{1}{2x}+\frac{7}{48}x+\dots \right]dx \\ =y_1\left(x\right)\left[-\frac{1}{2x}-\frac{1}{2}\ln x+\frac{7}{96}{x}^2+\dots \right] \end{gathered}$$

We can obtain a general solution by using Superposition Principle as follows

$\left.y=C_1\left[1+\frac{1}{4}{x}^2+\right.\frac{1}{48}{x}^4+\dots \right]+C_2\left[1+\frac{1}{4}{x}^2+\frac{1}{48}{x}^4+\dots \right]\left[-\frac{1}{2x}-\frac{1}{2}\ln x+\frac{7}{96}{x}^2+\dots \right]$