2.4.69 Problem 66

Internal problem ID [8958]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 66
Date solved : Wednesday, March 05, 2025 at 07:09:40 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

Using series expansion around $x=0$

Solving ode using Taylor series method. This gives review on how the Taylor series method works for solving second order ode.

Let

$$y^{\prime \prime }=f(x,y,y^{\prime })$$

Assuming expansion is at $x_0=0$ (we can always shift the actual expansion point to $0$ by change of variables) and assuming $f(x,y,y^{\prime })$ is analytic at $x_0$ which must be the case for an ordinary point. Let initial conditions be $y\left(x_0\right)=y_0$ and $y^{\prime }\left(x_0\right)=y_0^{\prime }$. Using Taylor series gives

$$\begin{array}{rl}y\left(x\right)& =y\left(x_0\right)+(x-x_0)y^{\prime }\left(x_0\right)+\frac{{(x-x_0)}^2}{2}{y}^{\prime \prime }\left(x_0\right)+\frac{{(x-x_0)}^3}{3!}{y}^{\prime \prime \prime }\left(x_0\right)+\cdots \\ & =y_0+x{y}_0^{\prime }+\frac{x^2}{2}{f|}_{x_0,y_0,y_0^{\prime }}+\frac{x^3}{3!}{f^{\prime }|}_{x_0,y_0,y_0^{\prime }}+\cdots \\ & =y_0+x{y}_0^{\prime }+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+2}}{(n+2)!}{\frac{d^{n}f}{d{x}^n}|}_{x_0,y_0,y_0^{\prime }}\end{array}$$

But

$$\begin{array}{rl}\text{(1)}& \frac{df}{dx}& =\frac{\partial f}{\partial x}\frac{dx}{dx}+\frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial y^{\prime }}\frac{d{y}^{\prime }}{dx}\text{(1)}& & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}{y}^{\prime }+\frac{\partial f}{\partial y^{\prime }}{y}^{\prime \prime }\text{(2)}& & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}{y}^{\prime }+\frac{\partial f}{\partial y^{\prime }}f\frac{d^{2}f}{d{x}^2}& =\frac{d}{dx}\left(\frac{df}{dx}\right)\\ \text{(2)}& & =\frac{\partial }{\partial x}\left(\frac{df}{dx}\right)+\frac{\partial }{\partial y}\left(\frac{df}{dx}\right)y^{\prime }+\frac{\partial }{\partial y^{\prime }}\left(\frac{df}{dx}\right)f\frac{d^{3}f}{d{x}^3}& =\frac{d}{dx}\left(\frac{d^{2}f}{d{x}^2}\right)\\ \text{(3)}& & =\frac{\partial }{\partial x}\left(\frac{d^{2}f}{d{x}^2}\right)+\left(\frac{\partial }{\partial y}\frac{d^{2}f}{d{x}^2}\right)y^{\prime }+\frac{\partial }{\partial y^{\prime }}\left(\frac{d^{2}f}{d{x}^2}\right)f& ⋮\end{array}$$

And so on. Hence if we name $F_0=f(x,y,y^{\prime })$ then the above can be written as

$$\begin{array}{rl}\text{(4)}& F_0& =f(x,y,y^{\prime })F_1& =\frac{df}{dx}\\ & =\frac{d{F}_0}{dx}\\ & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}{y}^{\prime }+\frac{\partial f}{\partial y^{\prime }}{y}^{\prime \prime }\\ \text{(5)}& & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}{y}^{\prime }+\frac{\partial f}{\partial y^{\prime }}f& =\frac{\partial F_0}{\partial x}+\frac{\partial F_0}{\partial y}{y}^{\prime }+\frac{\partial F_0}{\partial y^{\prime }}{F}_0\\ F_2& =\frac{d}{dx}\left(\frac{d}{dx}f\right)\\ & =\frac{d}{dx}\left(F_1\right)\\ & =\frac{\partial }{\partial x}{F}_1+\left(\frac{\partial F_1}{\partial y}\right)y^{\prime }+\left(\frac{\partial F_1}{\partial y^{\prime }}\right)y^{\prime \prime }\\ & =\frac{\partial }{\partial x}{F}_1+\left(\frac{\partial F_1}{\partial y}\right)y^{\prime }+\left(\frac{\partial F_1}{\partial y^{\prime }}\right)F_0\\ & ⋮\\ F_n& =\frac{d}{dx}\left(F_{n-1}\right)\\ & =\frac{\partial }{\partial x}{F}_{n-1}+\left(\frac{\partial F_{n-1}}{\partial y}\right)y^{\prime }+\left(\frac{\partial F_{n-1}}{\partial y^{\prime }}\right)y^{\prime \prime }\\ \text{(6)}& & =\frac{\partial }{\partial x}{F}_{n-1}+\left(\frac{\partial F_{n-1}}{\partial y}\right)y^{\prime }+\left(\frac{\partial F_{n-1}}{\partial y^{\prime }}\right)F_0\end{array}$$

Therefore (6) can be used from now on along with

$$\begin{array}{}\text{(7)}& y\left(x\right)=y_0+x{y}_0^{\prime }+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+2}}{(n+2)!}{F_n|}_{x_0,y_0,y_0^{\prime }}\end{array}$$

To find $y\left(x\right)$ series solution around $x=0$. Hence

$$\begin{array}{rl}{F}_0& =-(x-1)y\\ F_1& =\frac{d{F}_0}{dx}\\ & =\frac{\partial F_0}{\partial x}+\frac{\partial F_0}{\partial y}{y}^{\prime }+\frac{\partial F_0}{\partial y^{\prime }}{F}_0\\ & =-y-(x-1)y^{\prime }\\ F_2& =\frac{d{F}_1}{dx}\\ & =\frac{\partial F_1}{\partial x}+\frac{\partial F_1}{\partial y}{y}^{\prime }+\frac{\partial F_1}{\partial y^{\prime }}{F}_1\\ & =-2y^{\prime }+{(x-1)}^{2}y\\ F_3& =\frac{d{F}_2}{dx}\\ & =\frac{\partial F_2}{\partial x}+\frac{\partial F_2}{\partial y}{y}^{\prime }+\frac{\partial F_2}{\partial y^{\prime }}{F}_2\\ & =(x-1)((x-1)y^{\prime }+4y)\\ F_4& =\frac{d{F}_3}{dx}\\ & =\frac{\partial F_3}{\partial x}+\frac{\partial F_3}{\partial y}{y}^{\prime }+\frac{\partial F_3}{\partial y^{\prime }}{F}_3\\ & =-{(x-1)}^{3}y+(6x-6)y^{\prime }+4y\end{array}$$

And so on. Evaluating all the above at initial conditions $x=0$ and $y\left(0\right)=y\left(0\right)$ and $y^{\prime }\left(0\right)=y^{\prime }\left(0\right)$ gives

$$\begin{array}{rl}{F}_0& =y\left(0\right)\\ F_1& =-y\left(0\right)+y^{\prime }\left(0\right)\\ F_2& =-2y^{\prime }\left(0\right)+y\left(0\right)\\ F_3& =y^{\prime }\left(0\right)-4y\left(0\right)\\ F_4& =5y\left(0\right)-6y^{\prime }\left(0\right)\end{array}$$

Substituting all the above in (7) and simplifying gives the solution as

$$y=(1+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\frac{1}{24}{x}^4-\frac{1}{30}{x}^5)y\left(0\right)+(x+\frac{1}{6}{x}^3-\frac{1}{12}{x}^4+\frac{1}{120}{x}^5)y^{\prime }\left(0\right)+O\left(x^6\right)$$

Since the expansion point $x=0$ is an ordinary point, then this can also be solved using the standard power series method. Let the solution be represented as power series of the form

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

Then

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

Substituting the above back into the ode gives

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

Which simplifies to

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

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

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

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

$$\begin{array}{}\text{(3)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+2)a_{n+2}(1+n)x^n)+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_{n-1}{x}^n)+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-a_n{x}^n)=0\end{array}$$

$n=0$ gives

$$2a_2-a_0=0$$

$$a_2=\frac{a_0}{2}$$

For $1\le n$, the recurrence equation is

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

Solving for $a_{n+2}$, gives

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

For $n=1$ the recurrence equation gives

$$6a_3+a_0-a_1=0$$

Which after substituting the earlier terms found becomes

$$a_3=-\frac{a_0}{6}+\frac{a_1}{6}$$

For $n=2$ the recurrence equation gives

$$12a_4+a_1-a_2=0$$

Which after substituting the earlier terms found becomes

$$a_4=-\frac{a_1}{12}+\frac{a_0}{24}$$

For $n=3$ the recurrence equation gives

$$20a_5+a_2-a_3=0$$

Which after substituting the earlier terms found becomes

$$a_5=-\frac{a_0}{30}+\frac{a_1}{120}$$

For $n=4$ the recurrence equation gives

$$30a_6+a_3-a_4=0$$

Which after substituting the earlier terms found becomes

$$a_6=\frac{a_0}{144}-\frac{a_1}{120}$$

For $n=5$ the recurrence equation gives

$$42a_7+a_4-a_5=0$$

Which after substituting the earlier terms found becomes

$$a_7=\frac{11a_1}{5040}-\frac{a_0}{560}$$

And so on. Therefore the solution is

$$\begin{array}{rl}y& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^n\\ & =a_3{x}^3+a_2{x}^2+a_{1}x+a_0+\dots \end{array}$$

Substituting the values for $a_n$ found above, the solution becomes

$$y=a_0+a_{1}x+\frac{a_0{x}^2}{2}+(-\frac{a_0}{6}+\frac{a_1}{6})x^3+(-\frac{a_1}{12}+\frac{a_0}{24})x^4+(-\frac{a_0}{30}+\frac{a_1}{120})x^5+\dots$$

Collecting terms, the solution becomes

$$\begin{array}{}\text{(3)}& y=(1+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\frac{1}{24}{x}^4-\frac{1}{30}{x}^5)a_0+(x+\frac{1}{6}{x}^3-\frac{1}{12}{x}^4+\frac{1}{120}{x}^5)a_1+O\left(x^6\right)\end{array}$$

At $x=0$ the solution above becomes

$$y=(1+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\frac{1}{24}{x}^4-\frac{1}{30}{x}^5)c_1+(x+\frac{1}{6}{x}^3-\frac{1}{12}{x}^4+\frac{1}{120}{x}^5)c_2+O\left(x^6\right)$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 54

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

$$y=(1+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\frac{1}{24}{x}^4-\frac{1}{30}{x}^5)y\left(0\right)+(x+\frac{1}{6}{x}^3-\frac{1}{12}{x}^4+\frac{1}{120}{x}^5)y^{\prime }\left(0\right)+O\left(x^6\right)$$

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 
<- No Liouvillian solutions exists 
-> Trying a solution in terms of special functions: 
   -> Bessel 
   <- Bessel successful 
<- special function solution successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)+(x-1)y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{x}^m\cdot y\left(x\right)\text{to series expansion for}m=0..1\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,-m)}}{a}_k{x}^{k+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,-m)+m}}{a}_{k-m}{x}^k\\ & \circ & \text{Convert}\frac{d^2}{d{x}^2}y\left(x\right)\text{to series expansion}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=2}}{a}_{k}k(k-1)x^{k-2}\\ & \circ & \text{Shift index using}k->k+2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_{k+2}(k+2)(k+1)x^k\\ & & \text{Rewrite ODE with series expansions}\\ & & 2a_2-a_0+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_{k+2}(k+2)(k+1)-a_k+a_{k-1})x^k)=0\\ \bullet & & \text{Each term must be 0}\\ & & 2a_2-a_0=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k^2+3k+2)a_{k+2}+a_{k-1}-a_k=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & ({(k+1)}^2+3k+5)a_{k+3}+a_k-a_{k+1}=0\\ \bullet & & \text{Recursion relation that defines the series solution to the ODE}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+3}=-\frac{a_k-a_{k+1}}{k^2+5k+6},2a_2-a_0=0]\end{array}$$

✓ Mathematica. Time used: 0.001 (sec). Leaf size: 63

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

$$y(x)\to c_2(\frac{x^5}{120}-\frac{x^4}{12}+\frac{x^3}{6}+x)+c_1(-\frac{x^5}{30}+\frac{x^4}{24}-\frac{x^3}{6}+\frac{x^2}{2}+1)$$

✓ Sympy. Time used: 0.780 (sec). Leaf size: 39

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

$$y\left(x\right)=C_2(\frac{x^4}{24}-\frac{x^3}{6}+\frac{x^2}{2}+1)+C_{1}x(-\frac{x^3}{12}+\frac{x^2}{6}+1)+O\left(x^6\right)$$