2.4.38 Problem 35

Internal problem ID [8926]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 35
Date solved : Friday, April 25, 2025 at 05:24:47 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}x{y}^{\prime \prime }+(1+x)y^{\prime }+2y& =0\end{array}$$

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.

$$x{y}^{\prime \prime }+(1+x)y^{\prime }+2y=0$$

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

$$\begin{array}{rl}p(x)& =\frac{1+x}{x}\\ q(x)& =\frac{2}{x}\end{array}$$

Table 2.90: Table $p(x),q(x)$ singularites.

$p(x)=\frac{1+x}{x}$
singularity	type
$x=0$	$\text{“regular”}$

$q(x)=\frac{2}{x}$
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

$$x{y}^{\prime \prime }+(1+x)y^{\prime }+2y=0$$

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

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

Which simplifies to

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

The next step is to make all powers of $x$ be $n+r-1$ in each summation term. Going over each summation term above with power of $x$ in it which is not already $x^{n+r-1}$ and adjusting the power and the corresponding index gives

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

Substituting all the above in Eq (2A) gives the following equation where now all powers of $x$ are the same and equal to $n+r-1$.

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

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

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

When $n=0$ the above becomes

$$x^{-1+r}{a}_{0}r(-1+r)+r{a}_0{x}^{-1+r}=0$$

Or

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

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

$$x^{-1+r}{r}^2=0$$

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

$$r^2=0$$

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

$$\begin{array}{rl}{r}_1& =0\\ r_2& =0\end{array}$$

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

$$x^{-1+r}{r}^2=0$$

Solving for $r$ gives the roots of the indicial equation as $[0,0]$.

Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form

$$\begin{array}{}\text{(1A)}& y_1\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r}\end{array}$$

Now the second solution $y_2$ is found using

$$\begin{array}{}\text{(1B)}& y_2\left(x\right)& =y_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}{b}_n{x}^{n+r})\end{array}$$

Then the general solution will be

$$y=c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)$$

In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), $a_0$ is never zero, and is arbitrary and is typically taken as $a_0=1$, and $\{c_1,c_2\}$ are two arbitray constants of integration which can be found from initial conditions. We start by finding the first solution $y_1\left(x\right)$. 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 $a_0=1$. For $1\le n$ the recursive equation is

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

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

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

Which for the root $r=0$ becomes

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

At this point, it is a good idea to keep track of $a_n$ in a table both before substituting $r=0$ and after as more terms are found using the above recursive equation.

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

For $n=1$, using the above recursive equation gives

$$a_1=\frac{-2-r}{{(r+1)}^2}$$

Which for the root $r=0$ becomes

$$a_1=-2$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{{(r+1)}^2}$	$-2$

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

$$a_2=\frac{3+r}{(2+r){(r+1)}^2}$$

Which for the root $r=0$ becomes

$$a_2=\frac{3}{2}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{{(r+1)}^2}$	$-2$
$a_2$	$\frac{3+r}{(2+r){(r+1)}^2}$	$\frac{3}{2}$

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

$$a_3=\frac{-4-r}{(2+r)(3+r){(r+1)}^2}$$

Which for the root $r=0$ becomes

$$a_3=-\frac{2}{3}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{{(r+1)}^2}$	$-2$
$a_2$	$\frac{3+r}{(2+r){(r+1)}^2}$	$\frac{3}{2}$
$a_3$	$\frac{-4-r}{(2+r)(3+r){(r+1)}^2}$	$-\frac{2}{3}$

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

$$a_4=\frac{5+r}{(4+r)(2+r)(3+r){(r+1)}^2}$$

Which for the root $r=0$ becomes

$$a_4=\frac{5}{24}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{{(r+1)}^2}$	$-2$
$a_2$	$\frac{3+r}{(2+r){(r+1)}^2}$	$\frac{3}{2}$
$a_3$	$\frac{-4-r}{(2+r)(3+r){(r+1)}^2}$	$-\frac{2}{3}$
$a_4$	$\frac{5+r}{(4+r)(2+r)(3+r){(r+1)}^2}$	$\frac{5}{24}$

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

$$a_5=\frac{-6-r}{(5+r)(4+r)(2+r)(3+r){(r+1)}^2}$$

Which for the root $r=0$ becomes

$$a_5=-\frac{1}{20}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{{(r+1)}^2}$	$-2$
$a_2$	$\frac{3+r}{(2+r){(r+1)}^2}$	$\frac{3}{2}$
$a_3$	$\frac{-4-r}{(2+r)(3+r){(r+1)}^2}$	$-\frac{2}{3}$
$a_4$	$\frac{5+r}{(4+r)(2+r)(3+r){(r+1)}^2}$	$\frac{5}{24}$
$a_5$	$\frac{-6-r}{(5+r)(4+r)(2+r)(3+r){(r+1)}^2}$	$-\frac{1}{20}$

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

$$\begin{array}{rl}{y}_1\left(x\right)& =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots \\ & =1-2x+\frac{3x^2}{2}-\frac{2x^3}{3}+\frac{5x^4}{24}-\frac{x^5}{20}+O\left(x^6\right)\end{array}$$

Now the second solution is found. The second solution is given by

$$y_2\left(x\right)=y_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}{b}_n{x}^{n+r})$$

Where $b_n$ is found using

$$b_n=\frac{d}{dr}{a}_{n,r}$$

And the above is then evaluated at $r=0$. The above table for $a_{n,r}$ is used for this purpose. Computing the derivatives gives the following table

$n$	$b_{n,r}$	$a_n$	$b_{n,r}=\frac{d}{dr}{a}_{n,r}$	$b_n(r=0)$
$b_0$	$1$	$1$	N/A since $b_n$ starts from 1	N/A
$b_1$	$\frac{-2-r}{{(r+1)}^2}$	$-2$	$\frac{3+r}{{(r+1)}^3}$	$3$
$b_2$	$\frac{3+r}{(2+r){(r+1)}^2}$	$\frac{3}{2}$	$\frac{-2r^2-11r-13}{{(2+r)}^2{(r+1)}^3}$	$-\frac{13}{4}$
$b_3$	$\frac{-4-r}{(2+r)(3+r){(r+1)}^2}$	$-\frac{2}{3}$	$\frac{3r^3+27r^2+74r+62}{{(2+r)}^2{(3+r)}^2{(r+1)}^3}$	$\frac{31}{18}$
$b_4$	$\frac{5+r}{(4+r)(2+r)(3+r){(r+1)}^2}$	$\frac{5}{24}$	$\frac{-4r^4-54r^3-256r^2-504r-346}{{(4+r)}^2{(2+r)}^2{(3+r)}^2{(r+1)}^3}$	$-\frac{173}{288}$
$b_5$	$\frac{-6-r}{(5+r)(4+r)(2+r)(3+r){(r+1)}^2}$	$-\frac{1}{20}$	$\frac{5r^5+95r^4+685r^3+2335r^2+3744r+2244}{{(5+r)}^2{(4+r)}^2{(2+r)}^2{(3+r)}^2{(r+1)}^3}$	$\frac{187}{1200}$

The above table gives all values of $b_n$ needed. Hence the second solution is

$$\begin{array}{rl}{y}_2\left(x\right)& =y_1\left(x\right)\ln \left(x\right)+b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4+b_5{x}^5+b_6{x}^6\dots \\ & =(1-2x+\frac{3x^2}{2}-\frac{2x^3}{3}+\frac{5x^4}{24}-\frac{x^5}{20}+O\left(x^6\right))\ln \left(x\right)+3x-\frac{13x^2}{4}+\frac{31x^3}{18}-\frac{173x^4}{288}+\frac{187x^5}{1200}+O\left(x^6\right)\end{array}$$

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =c_1(1-2x+\frac{3x^2}{2}-\frac{2x^3}{3}+\frac{5x^4}{24}-\frac{x^5}{20}+O\left(x^6\right))+c_2((1-2x+\frac{3x^2}{2}-\frac{2x^3}{3}+\frac{5x^4}{24}-\frac{x^5}{20}+O\left(x^6\right))\ln \left(x\right)+3x-\frac{13x^2}{4}+\frac{31x^3}{18}-\frac{173x^4}{288}+\frac{187x^5}{1200}+O\left(x^6\right))\end{array}$$

Hence the final solution is

$$\begin{array}{rl}y& =y_h\\ & =c_1(1-2x+\frac{3x^2}{2}-\frac{2x^3}{3}+\frac{5x^4}{24}-\frac{x^5}{20}+O\left(x^6\right))+c_2((1-2x+\frac{3x^2}{2}-\frac{2x^3}{3}+\frac{5x^4}{24}-\frac{x^5}{20}+O\left(x^6\right))\ln \left(x\right)+3x-\frac{13x^2}{4}+\frac{31x^3}{18}-\frac{173x^4}{288}+\frac{187x^5}{1200}+O\left(x^6\right))\end{array}$$

✓ Maple. Time used: 0.029 (sec). Leaf size: 44

Order:=6; 
ode:=x*diff(diff(y(x),x),x)+(1+x)*diff(y(x),x)+2*y(x) = 0; 
dsolve(ode,y(x),type='series',x=0);
 

$$y=(c_2\ln \left(x\right)+c_1)(1-2x+\frac{3}{2}{x}^2-\frac{2}{3}{x}^3+\frac{5}{24}{x}^4-\frac{1}{20}{x}^5+\mathrm{O}\left(x^6\right))+(3x-\frac{13}{4}{x}^2+\frac{31}{18}{x}^3-\frac{173}{288}{x}^4+\frac{187}{1200}{x}^5+\mathrm{O}\left(x^6\right))c_2$$

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 
   Reducible group (found an exponential solution) 
   Group is reducible, not completely reducible 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(1+x)\left(\frac{d}{dx}y\left(x\right)\right)+2y\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{2y\left(x\right)}{x}-\frac{(1+x)\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{(1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x}+\frac{2y\left(x\right)}{x}=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{1+x}{x},P_3\left(x\right)=\frac{2}{x}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=1\\ & \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}=0\\ & \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\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(1+x)\left(\frac{d}{dx}y\left(x\right)\right)+2y\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 \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=0..1\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)x^{k+r}\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\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-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)(k+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_0{r}^2{x}^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(a_{k+1}{(k+1+r)}^2+a_k(k+r+2))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r^2=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}{(k+1)}^2+a_k(k+2)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=-\frac{a_k(k+2)}{{(k+1)}^2}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=-\frac{a_k(k+2)}{{(k+1)}^2}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+1}=-\frac{a_k(k+2)}{{(k+1)}^2}]\end{array}$$

✓ Mathematica. Time used: 0.007 (sec). Leaf size: 111

ode=x*D[y[x],{x,2}] +(1+x)*D[y[x],x]+2*y[x] == 0; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

$$y(x)\to c_1(-\frac{x^5}{20}+\frac{5x^4}{24}-\frac{2x^3}{3}+\frac{3x^2}{2}-2x+1)+c_2(\frac{187x^5}{1200}-\frac{173x^4}{288}+\frac{31x^3}{18}-\frac{13x^2}{4}+(-\frac{x^5}{20}+\frac{5x^4}{24}-\frac{2x^3}{3}+\frac{3x^2}{2}-2x+1)\log (x)+3x)$$

✓ Sympy. Time used: 0.823 (sec). Leaf size: 37

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x*Derivative(y(x), (x, 2)) + (x + 1)*Derivative(y(x), x) + 2*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
 

$$y\left(x\right)=C_1(-\frac{x^5}{20}+\frac{5x^4}{24}-\frac{2x^3}{3}+\frac{3x^2}{2}-2x+1)+O\left(x^6\right)$$