2.4.48 Problem 45

Internal problem ID [8937]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 45
Date solved : Sunday, March 30, 2025 at 01:55:50 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{x}^2{y}^{\prime \prime }+4x{y}^{\prime }+(x^2+2)y& =0\end{array}$$

Using series expansion around $x=0$

The type of the expansion point is first determined. This is done on the homogeneous part of the ODE.

$$x^2{y}^{\prime \prime }+4x{y}^{\prime }+(x^2+2)y=0$$

The following is summary of singularities for the above ode. Writing the ode as

$$\begin{array}{rl}{y}^{\prime \prime }+p(x)y^{\prime }+q(x)y& =0\end{array}$$

Where

$$\begin{array}{rl}p(x)& =\frac{4}{x}\\ q(x)& =\frac{x^2+2}{x^2}\end{array}$$

Table 2.100: Table $p(x),q(x)$ singularites.

$p(x)=\frac{4}{x}$
singularity	type
$x=0$	$\text{“regular”}$

$q(x)=\frac{x^2+2}{x^2}$
singularity	type
$x=0$	$\text{“regular”}$

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : $[0]$

Irregular singular points : $[\mathrm{\infty }]$

Since $x=0$ is regular singular point, then Frobenius power series is used. The ode is normalized to be

$$x^2{y}^{\prime \prime }+4x{y}^{\prime }+(x^2+2)y=0$$

Let the solution be represented as Frobenius power series of the form

$$y=\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r}$$

Then

$$\begin{array}{rl}{y}^{\prime }& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)a_n{x}^{n+r-1}\\ y^{\prime \prime }& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2}\end{array}$$

Substituting the above back into the ode gives

$$\begin{array}{}\text{(1)}& x^2(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2})+4x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)a_n{x}^{n+r-1})+(x^2+2)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r})=0\end{array}$$

Which simplifies to

$$\begin{array}{}\text{(2A)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}4x^{n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r+2}{a}_n)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}2a_n{x}^{n+r})=0\end{array}$$

The next step is to make all powers of $x$ be $n+r$ in each summation term. Going over each summation term above with power of $x$ in it which is not already $x^{n+r}$ and adjusting the power and the corresponding index gives

$$\begin{array}{rl}\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r+2}{a}_n& =\stackrel{\mathrm{\infty }}{\sum _{n=2}}{a}_{n-2}{x}^{n+r}\end{array}$$

Substituting all the above in Eq (2A) gives the following equation where now all powers of $x$ are the same and equal to $n+r$.

$$\begin{array}{}\text{(2B)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}4x^{n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=2}}{a}_{n-2}{x}^{n+r})+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}2a_n{x}^{n+r})=0\end{array}$$

The indicial equation is obtained from $n=0$. From Eq (2B) this gives

$$x^{n+r}{a}_n(n+r)(n+r-1)+4x^{n+r}{a}_n(n+r)+2a_n{x}^{n+r}=0$$

When $n=0$ the above becomes

$$x^r{a}_{0}r(-1+r)+4x^r{a}_{0}r+2a_0{x}^r=0$$

Or

$$(x^{r}r(-1+r)+4x^{r}r+2x^r)a_0=0$$

Since $a_0\ne 0$ then the above simplifies to

$$(r^2+3r+2)x^r=0$$

Since the above is true for all $x$ then the indicial equation becomes

$$r^2+3r+2=0$$

Solving for $r$ gives the roots of the indicial equation as

$$\begin{array}{rl}{r}_1& =-1\\ r_2& =-2\end{array}$$

Since $a_0\ne 0$ then the indicial equation becomes

$$(r^2+3r+2)x^r=0$$

Solving for $r$ gives the roots of the indicial equation as $[-1,-2]$.

Since $r_1-r_2=1$ is an integer, then we can construct two linearly independent solutions

$$\begin{array}{rl}{y}_1\left(x\right)& =x^{r_1}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^n)\\ y_2\left(x\right)& =C{y}_1\left(x\right)\ln \left(x\right)+x^{r_2}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)\end{array}$$

Or

$$\begin{array}{rl}{y}_1\left(x\right)& =\frac{\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^n}{x}\\ y_2\left(x\right)& =C{y}_1\left(x\right)\ln \left(x\right)+\frac{\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n}{x^2}\end{array}$$

Or

$$\begin{array}{rl}{y}_1\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n-1}\\ y_2\left(x\right)& =C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n-2})\end{array}$$

Where $C$ above can be zero. We start by finding $y_1$. Eq (2B) derived above is now used to find all $a_n$ coefficients. The case $n=0$ is skipped since it was used to find the roots of the indicial equation. $a_0$ is arbitrary and taken as $a_0=1$. Substituting $n=1$ in Eq. (2B) gives

$$a_1=0$$

For $2\le n$ the recursive equation is

$$\begin{array}{}\text{(3)}& a_n(n+r)(n+r-1)+4a_n(n+r)+a_{n-2}+2a_n=0\end{array}$$

Solving for $a_n$ from recursive equation (4) gives

$$\begin{array}{}\text{(4)}& a_n=-\frac{a_{n-2}}{n^2+2nr+r^2+3n+3r+2}\end{array}$$

Which for the root $r=-1$ becomes

$$\begin{array}{}\text{(5)}& a_n=-\frac{a_{n-2}}{n(n+1)}\end{array}$$

At this point, it is a good idea to keep track of $a_n$ in a table both before substituting $r=-1$ and after as more terms are found using the above recursive equation.

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$

For $n=2$, using the above recursive equation gives

$$a_2=-\frac{1}{r^2+7r+12}$$

Which for the root $r=-1$ becomes

$$a_2=-\frac{1}{6}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$-\frac{1}{r^2+7r+12}$	$-\frac{1}{6}$

For $n=3$, using the above recursive equation gives

$$a_3=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$-\frac{1}{r^2+7r+12}$	$-\frac{1}{6}$
$a_3$	$0$	$0$

For $n=4$, using the above recursive equation gives

$$a_4=\frac{1}{(r+4)(r+3)(r+6)(r+5)}$$

Which for the root $r=-1$ becomes

$$a_4=\frac{1}{120}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$-\frac{1}{r^2+7r+12}$	$-\frac{1}{6}$
$a_3$	$0$	$0$
$a_4$	$\frac{1}{(r+4)(r+3)(r+6)(r+5)}$	$\frac{1}{120}$

For $n=5$, using the above recursive equation gives

$$a_5=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$-\frac{1}{r^2+7r+12}$	$-\frac{1}{6}$
$a_3$	$0$	$0$
$a_4$	$\frac{1}{(r+4)(r+3)(r+6)(r+5)}$	$\frac{1}{120}$
$a_5$	$0$	$0$

Using the above table, then the solution $y_1\left(x\right)$ is

$$\begin{array}{rl}{y}_1\left(x\right)& =\frac{1}{x}(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots )\\ & =\frac{1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)}{x}\end{array}$$

Now the second solution $y_2\left(x\right)$ is found. Let

$$r_1-r_2=N$$

Where $N$ is positive integer which is the difference between the two roots. $r_1$ is taken as the larger root. Hence for this problem we have $N=1$. Now we need to determine if $C$ is zero or not. This is done by finding $\underset{r\to r_2}{lim}{a}_1\left(r\right)$. If this limit exists, then $C=0$, else we need to keep the log term and $C\ne 0$. The above table shows that

$$\begin{array}{rl}{a}_N& =a_1\\ & =0\end{array}$$

Therefore

$$\begin{array}{rl}\underset{r\to r_2}{lim}0& =\underset{r\to -2}{lim}0\\ & =0\end{array}$$

The limit is $0$. Since the limit exists then the log term is not needed and we can set $C=0$. Therefore the second solution has the form

$$\begin{array}{rl}{y}_2\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r}\\ & =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n-2}\end{array}$$

Eq (3) derived above is used to find all $b_n$ coefficients. The case $n=0$ is skipped since it was used to find the roots of the indicial equation. $b_0$ is arbitrary and taken as $b_0=1$. Substituting $n=1$ in Eq(3) gives

$$b_1=0$$

For $2\le n$ the recursive equation is

$$\begin{array}{}\text{(4)}& b_n(n+r)(n+r-1)+4b_n(n+r)+b_{n-2}+2b_n=0\end{array}$$

Which for for the root $r=-2$ becomes

$$\begin{array}{}\text{(4A)}& b_n(n-2)(n-3)+4b_n(n-2)+b_{n-2}+2b_n=0\end{array}$$

Solving for $b_n$ from the recursive equation (4) gives

$$\begin{array}{}\text{(5)}& b_n=-\frac{b_{n-2}}{n^2+2nr+r^2+3n+3r+2}\end{array}$$

Which for the root $r=-2$ becomes

$$\begin{array}{}\text{(6)}& b_n=-\frac{b_{n-2}}{n^2-n}\end{array}$$

At this point, it is a good idea to keep track of $b_n$ in a table both before substituting $r=-2$ and after as more terms are found using the above recursive equation.

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$

For $n=2$, using the above recursive equation gives

$$b_2=-\frac{1}{r^2+7r+12}$$

Which for the root $r=-2$ becomes

$$b_2=-\frac{1}{2}$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$-\frac{1}{r^2+7r+12}$	$-\frac{1}{2}$

For $n=3$, using the above recursive equation gives

$$b_3=0$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$-\frac{1}{r^2+7r+12}$	$-\frac{1}{2}$
$b_3$	$0$	$0$

For $n=4$, using the above recursive equation gives

$$b_4=\frac{1}{(r^2+7r+12)(r^2+11r+30)}$$

Which for the root $r=-2$ becomes

$$b_4=\frac{1}{24}$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$-\frac{1}{r^2+7r+12}$	$-\frac{1}{2}$
$b_3$	$0$	$0$
$b_4$	$\frac{1}{(r+4)(r+3)(r+6)(r+5)}$	$\frac{1}{24}$

For $n=5$, using the above recursive equation gives

$$b_5=0$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$-\frac{1}{r^2+7r+12}$	$-\frac{1}{2}$
$b_3$	$0$	$0$
$b_4$	$\frac{1}{(r+4)(r+3)(r+6)(r+5)}$	$\frac{1}{24}$
$b_5$	$0$	$0$

Using the above table, then the solution $y_2\left(x\right)$ is

$$\begin{array}{rl}{y}_2\left(x\right)& =\frac{1}{x}(b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4+b_5{x}^5+b_6{x}^6\dots )\\ & =\frac{1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)}{x^2}\end{array}$$

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =\frac{c_1(1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right))}{x}+\frac{c_2(1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right))}{x^2}\end{array}$$

Hence the final solution is

$$\begin{array}{rl}y& =y_h\\ & =\frac{c_1(1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right))}{x}+\frac{c_2(1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right))}{x^2}\end{array}$$

✓ Maple. Time used: 0.025 (sec). Leaf size: 34

Order:=6; 
ode:=x^2*diff(diff(y(x),x),x)+4*diff(y(x),x)*x+(x^2+2)*y(x) = 0; 
dsolve(ode,y(x),type='series',x=0);
 

$$y=\frac{c_1(1-\frac{1}{6}{x}^2+\frac{1}{120}{x}^4+\mathrm{O}\left(x^6\right))x+c_2(1-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4+\mathrm{O}\left(x^6\right))}{x^2}$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Group is reducible or imprimitive 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+4x\left(\frac{d}{dx}y\left(x\right)\right)+(x^2+2)y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=-\frac{(x^2+2)y\left(x\right)}{x^2}-\frac{4\left(\frac{d}{dx}y\left(x\right)\right)}{x}\\ \bullet & & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)+\frac{4\left(\frac{d}{dx}y\left(x\right)\right)}{x}+\frac{(x^2+2)y\left(x\right)}{x^2}=0\\ ◻& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{4}{x},P_3\left(x\right)=\frac{x^2+2}{x^2}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=4\\ & \circ & x^2\cdot P_3\left(x\right)\text{is analytic at}x=0\\ & & (x^2\cdot P_3\left(x\right))|{}_{x=0}=2\\ & \circ & x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+4x\left(\frac{d}{dx}y\left(x\right)\right)+(x^2+2)y\left(x\right)=0\\ \bullet & & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{x}^m\cdot y\left(x\right)\text{to series expansion for}m=0..2\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{x}^{k+r}\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r}\\ & \circ & \text{Convert}{x}^2\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x^2\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_0(2+r)(1+r)x^r+a_1(3+r)(2+r)x^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_k(k+r+2)(k+r+1)+a_{k-2})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & (2+r)(1+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-2,-1\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(3+r)(2+r)=0\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & a_1=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_k(k+r+2)(k+r+1)+a_{k-2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & a_{k+2}(k+4+r)(k+3+r)+a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{a_k}{(k+4+r)(k+3+r)}\\ \bullet & & \text{Recursion relation for}r=-2\\ & & a_{k+2}=-\frac{a_k}{(k+2)(k+1)}\\ \bullet & & \text{Solution for}r=-2\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-2},a_{k+2}=-\frac{a_k}{(k+2)(k+1)},a_1=0]\\ \bullet & & \text{Recursion relation for}r=-1\\ & & a_{k+2}=-\frac{a_k}{(k+3)(k+2)}\\ \bullet & & \text{Solution for}r=-1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-1},a_{k+2}=-\frac{a_k}{(k+3)(k+2)},a_1=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-2})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k-1}),a_{k+2}=-\frac{a_k}{(k+2)(k+1)},a_1=0,b_{k+2}=-\frac{b_k}{(k+3)(k+2)},b_1=0]\end{array}$$

✓ Mathematica. Time used: 0.013 (sec). Leaf size: 40

ode=x^2*D[y[x],{x,2}]+4*x*D[y[x],x]+(x^2+2)*y[x]==0; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

$$y(x)\to c_2(\frac{x^3}{120}-\frac{x}{6}+\frac{1}{x})+c_1(\frac{x^2}{24}+\frac{1}{x^2}-\frac{1}{2})$$

✓ Sympy. Time used: 0.849 (sec). Leaf size: 24

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*Derivative(y(x), (x, 2)) + 4*x*Derivative(y(x), x) + (x**2 + 2)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
 

$$y\left(x\right)=\frac{C_1(-\frac{x^6}{5040}+\frac{x^4}{120}-\frac{x^2}{6}+1)}{x}+O\left(1\right)$$