2.4.5 Problem 5

Internal problem ID [8894]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 5
Date solved : Wednesday, March 05, 2025 at 07:07:40 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solve

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

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.

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

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

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

Table 2.57: Table $p(x),q(x)$ singularites.

$p(x)=-\frac{1}{2x}$
singularity	type
$x=0$	$\text{“regular”}$

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

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

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

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

Which simplifies to

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

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 $n+r$.

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

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

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

When $n=0$ the above becomes

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

Or

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

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

$$(2r^2-3r+1)x^r=0$$

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

$$2r^2-3r+1=0$$

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

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

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

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

This equation will used later to find the particular solution.

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

$$(2r^2-3r+1)x^r=0$$

Solving for $r$ gives the roots of the indicial equation as $[1,\frac{1}{2}]$.

Since $r_1-r_2=\frac{1}{2}$ is not an integer, then we can construct two linearly independent solutions

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

Or

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

We start by finding $y_1\left(x\right)$. 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 $a_0=1$. Substituting $n=1$ in Eq. (2B) gives

$$a_1=0$$

For $2\le n$ the recursive equation is

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

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

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

Which for the root $r=1$ becomes

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

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

$$a_2=\frac{1}{2r^2+5r+3}$$

Which for the root $r=1$ becomes

$$a_2=\frac{1}{10}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$\frac{1}{2r^2+5r+3}$	$\frac{1}{10}$

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

$$a_3=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$\frac{1}{2r^2+5r+3}$	$\frac{1}{10}$
$a_3$	$0$	$0$

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

$$a_4=\frac{1}{4r^4+36r^3+113r^2+144r+63}$$

Which for the root $r=1$ becomes

$$a_4=\frac{1}{360}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$\frac{1}{2r^2+5r+3}$	$\frac{1}{10}$
$a_3$	$0$	$0$
$a_4$	$\frac{1}{4r^4+36r^3+113r^2+144r+63}$	$\frac{1}{360}$

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

$$a_5=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$\frac{1}{2r^2+5r+3}$	$\frac{1}{10}$
$a_3$	$0$	$0$
$a_4$	$\frac{1}{4r^4+36r^3+113r^2+144r+63}$	$\frac{1}{360}$
$a_5$	$0$	$0$

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

$$\begin{array}{rl}{y}_1\left(x\right)& =x(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots )\\ & =x(1+\frac{x^2}{10}+\frac{x^4}{360}+O\left(x^6\right))\end{array}$$

Now the second solution $y_2\left(x\right)$ is found. Eq (2B) derived above is now 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. (2B) gives

$$b_1=0$$

For $2\le n$ the recursive equation is

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

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

$$\begin{array}{}\text{(4)}& b_n=\frac{b_{n-2}}{2n^2+4nr+2r^2-3n-3r+1}\end{array}$$

Which for the root $r=\frac{1}{2}$ becomes

$$\begin{array}{}\text{(5)}& b_n=\frac{b_{n-2}}{n(2n-1)}\end{array}$$

At this point, it is a good idea to keep track of $b_n$ in a table both before substituting $r=\frac{1}{2}$ 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

$$b_2=\frac{1}{2r^2+5r+3}$$

Which for the root $r=\frac{1}{2}$ becomes

$$b_2=\frac{1}{6}$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$\frac{1}{2r^2+5r+3}$	$\frac{1}{6}$

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

$$b_3=0$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$\frac{1}{2r^2+5r+3}$	$\frac{1}{6}$
$b_3$	$0$	$0$

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

$$b_4=\frac{1}{4r^4+36r^3+113r^2+144r+63}$$

Which for the root $r=\frac{1}{2}$ becomes

$$b_4=\frac{1}{168}$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$\frac{1}{2r^2+5r+3}$	$\frac{1}{6}$
$b_3$	$0$	$0$
$b_4$	$\frac{1}{4r^4+36r^3+113r^2+144r+63}$	$\frac{1}{168}$

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

$$b_5=0$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$\frac{1}{2r^2+5r+3}$	$\frac{1}{6}$
$b_3$	$0$	$0$
$b_4$	$\frac{1}{4r^4+36r^3+113r^2+144r+63}$	$\frac{1}{168}$
$b_5$	$0$	$0$

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

$$\begin{array}{rl}{y}_2\left(x\right)& =x(b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4+b_5{x}^5+b_6{x}^6\dots )\\ & =\sqrt{x}(1+\frac{x^2}{6}+\frac{x^4}{168}+O\left(x^6\right))\end{array}$$

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =c_{1}x(1+\frac{x^2}{10}+\frac{x^4}{360}+O\left(x^6\right))+c_2\sqrt{x}(1+\frac{x^2}{6}+\frac{x^4}{168}+O\left(x^6\right))\end{array}$$

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

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

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 $r$.

$a_1=0$

$a_2=\frac{a_0}{2r^2+5r+3}$

$a_3=0$

$a_4=\frac{a_0}{(2r^2+5r+3)(2r^2+13r+21)}$

$a_5=0$

Since the $F=x^2+x+1$ has more than one term then we find a particular solution for each term and add the result to find the particular solution to the ode.

Now we determine the particular solution $y_p$ associated with $F=x^2$ by solving the balance equation

$$(2x^{m}m(-1+m)-x^{m}m+x^m)c_0=x^2$$

For $c_0$ and $x$. This results in

$$\begin{array}{rl}{c}_0& =\frac{1}{3}\\ m& =2\end{array}$$

The particular solution is therefore

$$\begin{array}{rl}{y}_p& =\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^{n+m}\\ & =\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^{n+2}\end{array}$$

Where in the above $c_0=\frac{1}{3}$.

The remaining $c_n$ values are found using the same recurrence relation given in the earlier table which was used to find the homogeneous solution but using $c_0$ in place of $a_0$ and using $m=2$ in place of the root of the indicial equation used to find the homogeneous solution. By letting $a_0=c_0$ or $a_0=\frac{1}{3}$ and $r=m$ or $r=2$. The following table gives the resulting $c_n$ values. These values will be used to find the particular solution. Values of $c_n$ found not defined when doing the substitution will be discarded and not used

$c_0=\frac{1}{3}$

$c_1=0$

$c_2=\frac{1}{63}$

$c_3=0$

$c_4=\frac{1}{3465}$

$c_5=0$

The particular solution is now found using

$$\begin{array}{rl}{y}_p& =x^m\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n\\ & =x^2\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n\end{array}$$

Using the values found above for $c_n$ into the above sum gives

$$\begin{array}{rl}{y}_p& =x^2(\frac{1}{3}+\frac{1}{63}{x}^2+\frac{1}{3465}{x}^4)\\ & =\frac{1}{3}{x}^2+\frac{1}{63}{x}^4+\frac{1}{3465}{x}^6\end{array}$$

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

Failed to convert RHS $x^2+x+1$ 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)-x*diff(y(x),x)+(-x^2+1)*y(x) = x^2+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: 0.046 (sec). Leaf size: 224

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

$$y(x)\to c_1\sqrt{x}(\frac{x^6}{11088}+\frac{x^4}{168}+\frac{x^2}{6}+1)+c_{2}x(\frac{x^6}{28080}+\frac{x^4}{360}+\frac{x^2}{10}+1)+\sqrt{x}(-\frac{79x^{11/2}}{154440}-\frac{x^{9/2}}{1620}-\frac{37x^{7/2}}{1260}-\frac{x^{5/2}}{25}-\frac{11x^{3/2}}{15}-2\sqrt{x}+\frac{2}{\sqrt{x}})(\frac{x^6}{11088}+\frac{x^4}{168}+\frac{x^2}{6}+1)+x(\frac{x^6}{28080}+\frac{x^4}{360}+\frac{x^2}{10}+1)(\frac{x^6}{66528}+\frac{67x^5}{55440}+\frac{x^4}{672}+\frac{29x^3}{504}+\frac{x^2}{12}+\frac{7x}{6}-\frac{1}{x}+\log (x))$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x**2*Derivative(y(x), (x, 2)) - x**2 - x*Derivative(y(x), x) - x + (1 - x**2)*y(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**2 - x*Derivative(y(x), x) - x + (1 - x**2)*y(x) - 1 does not match hint 2nd_power_series_regular