2.4.20 Problem 20

Internal problem ID [8908]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 20
Date solved : Friday, April 25, 2025 at 05:22:58 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solve

$$\begin{array}{rl}2x^2{y}^{\prime \prime }+2x{y}^{\prime }-xy& =1\end{array}$$

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.

$$2x^2{y}^{\prime \prime }+2x{y}^{\prime }-xy=0$$

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

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

Table 2.76: Table $p(x),q(x)$ singularites.

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

$q(x)=-\frac{1}{2x}$
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

$$2x^2{y}^{\prime \prime }+2x{y}^{\prime }-xy=1$$

Since this is an inhomogeneous, then let the solution be

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ode $2x^2{y}^{\prime \prime }+2x{y}^{\prime }-xy=0$, and $y_p$ is a particular solution to the inhomogeneous ode.which is found using the balance equation generated from indicial equation

First, we solve for $y_h$ 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

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

Which simplifies to

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

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

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

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

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

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

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

When $n=0$ the above becomes

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

Or

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

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

$$2x^r{r}^2=0$$

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

$$2r^2=0$$

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

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

The corresponding balance equation is found by replacing $r$ by $m$ and $a$ by $c$ to avoid confusing terms between particular solution and the homogeneous solution. Hence the balance equation is

$$\begin{array}{r}(2x^{m}m(-1+m)+2x^{m}m)c_0=1\end{array}$$

This equation will used later to find the particular solution.

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

$$2x^r{r}^2=0$$

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

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

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

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

Which for the root $r=0$ becomes

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

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

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

Which for the root $r=0$ becomes

$$a_1=\frac{1}{2}$$

And the table now becomes

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

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

$$a_2=\frac{1}{4{(r+1)}^2{(r+2)}^2}$$

Which for the root $r=0$ becomes

$$a_2=\frac{1}{16}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1}{2{(r+1)}^2}$	$\frac{1}{2}$
$a_2$	$\frac{1}{4{(r+1)}^2{(r+2)}^2}$	$\frac{1}{16}$

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

$$a_3=\frac{1}{8{(r+1)}^2{(r+2)}^2{(r+3)}^2}$$

Which for the root $r=0$ becomes

$$a_3=\frac{1}{288}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1}{2{(r+1)}^2}$	$\frac{1}{2}$
$a_2$	$\frac{1}{4{(r+1)}^2{(r+2)}^2}$	$\frac{1}{16}$
$a_3$	$\frac{1}{8{(r+1)}^2{(r+2)}^2{(r+3)}^2}$	$\frac{1}{288}$

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

$$a_4=\frac{1}{16{(r+1)}^2{(r+2)}^2{(r+3)}^2{(r+4)}^2}$$

Which for the root $r=0$ becomes

$$a_4=\frac{1}{9216}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1}{2{(r+1)}^2}$	$\frac{1}{2}$
$a_2$	$\frac{1}{4{(r+1)}^2{(r+2)}^2}$	$\frac{1}{16}$
$a_3$	$\frac{1}{8{(r+1)}^2{(r+2)}^2{(r+3)}^2}$	$\frac{1}{288}$
$a_4$	$\frac{1}{16{(r+1)}^2{(r+2)}^2{(r+3)}^2{(r+4)}^2}$	$\frac{1}{9216}$

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

$$a_5=\frac{1}{32{(r+1)}^2{(r+2)}^2{(r+3)}^2{(r+4)}^2{(r+5)}^2}$$

Which for the root $r=0$ becomes

$$a_5=\frac{1}{460800}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1}{2{(r+1)}^2}$	$\frac{1}{2}$
$a_2$	$\frac{1}{4{(r+1)}^2{(r+2)}^2}$	$\frac{1}{16}$
$a_3$	$\frac{1}{8{(r+1)}^2{(r+2)}^2{(r+3)}^2}$	$\frac{1}{288}$
$a_4$	$\frac{1}{16{(r+1)}^2{(r+2)}^2{(r+3)}^2{(r+4)}^2}$	$\frac{1}{9216}$
$a_5$	$\frac{1}{32{(r+1)}^2{(r+2)}^2{(r+3)}^2{(r+4)}^2{(r+5)}^2}$	$\frac{1}{460800}$

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

$$\begin{array}{rl}{y}_1\left(x\right)& =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots \\ & =1+\frac{x}{2}+\frac{x^2}{16}+\frac{x^3}{288}+\frac{x^4}{9216}+\frac{x^5}{460800}+O\left(x^6\right)\end{array}$$

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
$b_1$	$\frac{1}{2{(r+1)}^2}$	$\frac{1}{2}$	$-\frac{1}{{(r+1)}^3}$	$-1$
$b_2$	$\frac{1}{4{(r+1)}^2{(r+2)}^2}$	$\frac{1}{16}$	$\frac{-2r-3}{2{(r+1)}^3{(r+2)}^3}$	$-\frac{3}{16}$
$b_3$	$\frac{1}{8{(r+1)}^2{(r+2)}^2{(r+3)}^2}$	$\frac{1}{288}$	$\frac{-3r^2-12r-11}{4{(r+1)}^3{(r+2)}^3{(r+3)}^3}$	$-\frac{11}{864}$
$b_4$	$\frac{1}{16{(r+1)}^2{(r+2)}^2{(r+3)}^2{(r+4)}^2}$	$\frac{1}{9216}$	$\frac{-2r^3-15r^2-35r-25}{4{(r+1)}^3{(r+2)}^3{(r+3)}^3{(r+4)}^3}$	$-\frac{25}{55296}$
$b_5$	$\frac{1}{32{(r+1)}^2{(r+2)}^2{(r+3)}^2{(r+4)}^2{(r+5)}^2}$	$\frac{1}{460800}$	$\frac{-5r^4-60r^3-255r^2-450r-274}{16{(r+1)}^3{(r+2)}^3{(r+3)}^3{(r+4)}^3{(r+5)}^3}$	$-\frac{137}{13824000}$

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

$$\begin{array}{rl}{y}_2\left(x\right)& =y_1\left(x\right)\ln \left(x\right)+b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4+b_5{x}^5+b_6{x}^6\dots \\ & =(1+\frac{x}{2}+\frac{x^2}{16}+\frac{x^3}{288}+\frac{x^4}{9216}+\frac{x^5}{460800}+O\left(x^6\right))\ln \left(x\right)-x-\frac{3x^2}{16}-\frac{11x^3}{864}-\frac{25x^4}{55296}-\frac{137x^5}{13824000}+O\left(x^6\right)\end{array}$$

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =c_1(1+\frac{x}{2}+\frac{x^2}{16}+\frac{x^3}{288}+\frac{x^4}{9216}+\frac{x^5}{460800}+O\left(x^6\right))+c_2((1+\frac{x}{2}+\frac{x^2}{16}+\frac{x^3}{288}+\frac{x^4}{9216}+\frac{x^5}{460800}+O\left(x^6\right))\ln \left(x\right)-x-\frac{3x^2}{16}-\frac{11x^3}{864}-\frac{25x^4}{55296}-\frac{137x^5}{13824000}+O\left(x^6\right))\end{array}$$

The particular solution is found by solving for $c,m$ the balance equation

$$\begin{array}{rl}(2x^{m}m(-1+m)+2x^{m}m)c_0& =F\end{array}$$

Where $F(x)$ is the RHS of the ode. If $F(x)$ has more than one term, then this is done for each term one at a time and then all the particular solutions are added. The function $F(x)$ will be converted to series if needed. in order to solve for $c_n,m$ for each term, the same recursive relation used to find $y_h(x)$ is used to find $c_n,m$ which is used to find the particular solution $\sum _{n=0}{c}_n{x}^{n+m}$ by replacing $a_n$ by $c_n$ and $r$ by $m$.

The following are the values of $a_n$ found in terms of the indicial root $r$.

$a_1=\frac{a_0}{2{(r+1)}^2}$

$a_2=\frac{a_0}{4{(r+1)}^2{(r+2)}^2}$

$a_3=\frac{a_0}{8{(r+1)}^2{(r+2)}^2{(r+3)}^2}$

$a_4=\frac{a_0}{16{(r+1)}^2{(r+2)}^2{(r+3)}^2{(r+4)}^2}$

$a_5=\frac{a_0}{32{(r+1)}^2{(r+2)}^2{(r+3)}^2{(r+4)}^2{(r+5)}^2}$

Unable to solve the balance equation $(2x^{m}m(-1+m)+2x^{m}m)c_0$ for $c_0$ and $x$. No particular solution exists.

Failed to convert RHS $1$ to series in order to find particular solution. Unable to solve. Terminating Unable to find the particular solution or no solution exists.

✗ Maple

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

$$\text{No solution found}$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   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 
   <- No Liouvillian solutions exists 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      <- Bessel successful 
   <- special function solution successful 
<- solving first the homogeneous part of the ODE successful
 

✓ Mathematica. Time used: 0.144 (sec). Leaf size: 360

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

$$y(x)\to c_2(\frac{x^5}{460800}+\frac{x^4}{9216}+\frac{x^3}{288}+\frac{x^2}{16}+\frac{x}{2}+1)+c_1(x^5(\frac{\log (x)}{460800}-\frac{107}{13824000})+x^4(\frac{\log (x)}{9216}-\frac{19}{55296})+x^3(\frac{\log (x)}{288}-\frac{1}{108})+x^2(\frac{\log (x)}{16}-\frac{1}{8})+x(\frac{\log (x)}{2}-\frac{1}{2})+\log (x)+1)+(-\frac{137x^6}{1990656000}+\frac{x^5}{4608000}+\frac{x^4}{73728}+\frac{x^3}{1728}+\frac{x^2}{64}+\frac{x}{4}+\frac{\log (x)}{2})(x^5(\frac{\log (x)}{460800}-\frac{107}{13824000})+x^4(\frac{\log (x)}{9216}-\frac{19}{55296})+x^3(\frac{\log (x)}{288}-\frac{1}{108})+x^2(\frac{\log (x)}{16}-\frac{1}{8})+x(\frac{\log (x)}{2}-\frac{1}{2})+\log (x)+1)+(\frac{x^5}{460800}+\frac{x^4}{9216}+\frac{x^3}{288}+\frac{x^2}{16}+\frac{x}{2}+1)(\frac{137x^6(6\log (x)[phone]\mathrm{}}{\mathrm{}}+\frac{x^5(113-30\log (x))}{138240000}+\frac{x^4(41-12\log (x))}{884736}+\frac{x^3(3-\log (x))}{1728}+\frac{1}{128}{x}^2(5-2\log (x))+\frac{1}{4}x(2-\log (x))-\frac{1}{4}\log (x)(\log (x)+2))$$

✗ Sympy

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

ValueError : ODE 2*x**2*Derivative(y(x), (x, 2)) - x*y(x) + 2*x*Derivative(y(x), x) - 1 does not match hint 2nd_power_series_regular