Question

Each of the differential equations

${\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{"}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$ and ${\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{"}}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$

has an irregular singular point at $\mathit{x}\mathbf{=}\mathbf{0}$. Determine whether the method of Frobenius yields a Series solution of each differential equation about $\mathit{x}\mathbf{=}\mathbf{0}$.Discuss and explain your findings.

Short answer

For the first differential equation, the only solution is the trivial one, which is $y=0$.

For the second differential equation, it is the single series solution, which is$r=0$ .

Step-by-step solution

Definition of forbinous theorem

If$\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{0}}$is a regular singular point of the differential equation, then there exists at least one solution of the form

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{(}\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{0}}\mathbf{)}}^{\mathbf{r}}\sum _{n=0}^{\infty }{c}_n{(x-x_0)}^n \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\sum _{n=0}^{\infty }{c}_n{(x-x_0)}^{n+r} \end{gathered}$$

Where the number r is a constant to be determined.

The series will converge at least on some interval$\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{0}}\mathbf{<}\mathbf{R}$.

We have ${\mathrm{x}}^3{\mathrm{y}}^{"}+\mathrm{y}=0$and ${\mathrm{x}}^2{\mathrm{y}}^{"}+(3\mathrm{x}-1){\mathrm{y}}^{\text{'}}+\mathrm{y}=0$

which has an irregular singular point at$\mathrm{x}=0$

Use Forbinous methodto solve this differential equation

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

around$\mathrm{x}=0$

The first and second derivatives are:

$\mathrm{y}\text{'}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r}){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-1}$ …………..(2)

$\mathrm{y}"=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}$ ……………(3)

Step 2: Find the first differential equation

Substituting (1) and (3),

$f(y",y)=x^3\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-2}+\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

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

Now, make the summation index and the power of x, for the two series, in the same phase.

Assure that the first terms, for the two series, are raised to the same power, which is not the case.

$f(y",y)=x^3\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r+1}+c_0{x}^r+\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

Shifting the second series from$n$ to $n+1$,

$f(y",y)=\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r+1}+c_0{x}^r+\sum _{n+1=1}^{\infty }{c}_{n+1}{x}^{(n+1)+r}$

$=\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r+1}+c_0{x}^r+\sum _{n=0}^{\infty }{c}_{n+1}{x}^{n+r+1}$

Add thecoefficients of the terms, which are raised to the same power together,

$\begin{array}{rcl}f(y",y)& =& \sum _{n=0}^{\infty }[c_n(n+r)(n+r-1)+c_{n+1}]x^{n+r+1}+c_0{x}^r\\ & =& 0\end{array}$

which implies that

And also, $\begin{array}{l}{c}_0=0\\ c_1=c_2=c_3=\mathrm{....}=0\end{array}$

Therefore, we get the trivial solution which is y=0.

Step 3: Find the second differential equation

Substitute (1), (2) and (3) into the given differential equation,

$\mathrm{f}(\mathrm{y}",\mathrm{y}\text{'},\mathrm{y})={\mathrm{x}}^2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r})(\mathrm{n}+\mathrm{r}-1){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-2}+(3\mathrm{x}-1)\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}(\mathrm{n}+\mathrm{r}){\mathrm{x}}^{\mathrm{n}+\mathrm{r}-1}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+\mathrm{r}}$

$\Rightarrow \sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\infty }3c_n(n+r)x^{n+r}-\sum _{n=0}^{\infty }{c}_n(n+r)x^{n+r-1}+\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

Now, make the summation index and the power of x, for the four series, in the same phase.

Assure that the first terms, for the four series, are raised to the same power, which is not the case.

role="math" localid="1667969251459" $f(y",y\text{'},y)=\sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\infty }3c_n(n+r)x^{n+r}-[c_{0}r{x}^{r-1}+\sum _{n=1}^{\infty }{c}_n(n+r)x^{n+r-1}]+\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

$\Rightarrow \sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\infty }3c_n(n+r)x^{n+r}-c_{0}r{x}^{r-1}-\sum _{n=1}^{\infty }{c}_n(n+r)x^{n+r-1}+\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

Shifting the third series from $n$to $n+1$,

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

$\Rightarrow \sum _{n=0}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\infty }3c_n(n+r)x^{n+r}-c_{0}r{x}^{r-1}-\sum _{n=0}^{\infty }{c}_{n+1}(n+r+1)x^{n+r}+\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

Add the coefficients of the terms, which are raised to the same power together,

$f(y",y\text{'},y)=\sum _{n=0}^{\infty }[c_n(n+r)(n+r-1)+3c_n(n+r)-c_{n+1}(n+r+1)+c_n]x^{n+r}-c_{0}r{x}^{r-1}$

$\Rightarrow \sum _{n=0}^{\infty }[c_n(n+r)(n+r-1+3)-c_{n+1}(n+r+1)+c_n]x^{n+r}-c_{0}r{x}^{r-1}$

$\Rightarrow \sum _{n=0}^{\infty }\left[c_n\right[(n+r)(n+r+2)+1]-c_{n+1}(n+r+1)]x^{n+r}-c_{0}r{x}^{r-1}$

Which implies that

$c_n[(n+r)(n+r+2)+1]-c_{n+1}(n+r+1)=0$

$\Rightarrow c_{n+1}=\frac{c_n[(n+r)(n+r+2)+1]}{(n+r+1)}$ where, $\left[n=0,1,2,\mathrm{...}\right]$

We can’t set $c_0$to be zero as all the coefficients of the series are involved with $c_0$

At $r=0$, the solution has the form $y=\sum _{n=0}^{\infty }{c}_0{x}^n$

And the recurrence relation is ${\mathrm{c}}_{\mathrm{n}+1}=\frac{{\mathrm{c}}_{\mathrm{n}}[\mathrm{n}(\mathrm{n}+2)+1]}{(\mathrm{n}+1)}$

$\begin{array}{l}=\frac{{\mathrm{c}}_{\mathrm{n}}[{\mathrm{n}}^2+2\mathrm{n}+1]}{(\mathrm{n}+1)}\\ =\frac{{\mathrm{c}}_{\mathrm{n}}\left[{(\mathrm{n}+1)}^2\right]}{(\mathrm{n}+1)}\\ =(\mathrm{n}+1){\mathrm{c}}_{\mathrm{n}}\end{array}$

Now, find the coefficient of the series,

At , $\begin{array}{l}\mathrm{n}=0,{\mathrm{c}}_1={\mathrm{c}}_0\\ \mathrm{n}=1,{\mathrm{c}}_2=2{\mathrm{c}}_1=2{\mathrm{c}}_0\\ \mathrm{n}=2,{\mathrm{c}}_3=3{\mathrm{c}}_2=3\left(2{\mathrm{c}}_0\right)=6{\mathrm{c}}_0\\ \mathrm{n}=3,\mathrm{c}4=4{\mathrm{c}}_3=\mathrm{4.3.2}{\mathrm{c}}_0\end{array}$

…….. $c_n=n!c_0$ where,$\left[n=0,1,2,\mathrm{...}\right]$

And the serious solution is,

$\begin{array}{rcl}\mathrm{y}& =& \sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\\ & =& {\mathrm{c}}_0\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}!{\mathrm{x}}^{\mathrm{n}}\end{array}$

We apply the ratio test upon the above series,

$\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{(n+1)!x^{n+1}}{n!x^n}\right|$

$=\underset{n\to \infty }{\lim }\left|\frac{(n+1)\cancel{n!}{x}^{n+1}}{\cancel{n!}{x}^n}\right|$

$=\underset{n\to \infty }{\lim }\left|(n+1)x\right|$

since x is not anymore dependent on n and n is positive,

$\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\left|x\right|\underset{n\to \infty }{\lim }(n+1)$

which implies that the power series is convergent at the value of,upon which the limit dies, which is 0.

Using Forbinous method, obtain a single series solution for the second differential equation and it converges at$x=0$

Therefore,

For the first differential equation,$x^3{y}^{"}+y=0$the only solution is the trivial one, which is$y=0$.

And,for the second differential equation,$x^2{y}^{"}+(3x-1)y^{\text{'}}+y=0$it is the single series solution, which is$r=0$.