2.5.7 Problem 7

Internal problem ID [8968]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 7
Date solved : Sunday, March 30, 2025 at 01:57:33 PM
CAS classification : [[_linear, `class A`]]

Solve

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

Using series expansion around $x=0$

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 $y^{\prime }+y=0$,and $y_p$ is a particular solution to the inhomogeneous ode. 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

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

Substituting the above back into the ode gives

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

Which simplifies to

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

The next step is to make all powers of $x$ be $n+r-1$ 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

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

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

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

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

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

When $n=0$ the above becomes

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

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

$$\begin{array}{r}m{c}_0{x}^{-1+m}=\frac{1}{x}\end{array}$$

This equation will used later to find the particular solution.

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

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

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

$$r=0$$

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

$$r=0$$

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

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

For $n=1$ the recurrence equation gives

$$a_1(1+r)+a_0=0$$

Which after substituting the earlier terms found becomes

$$a_1=-\frac{a_0}{1+r}$$

For $n=2$ the recurrence equation gives

$$a_2(2+r)+a_1=0$$

Which after substituting the earlier terms found becomes

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

For $n=3$ the recurrence equation gives

$$a_3(3+r)+a_2=0$$

Which after substituting the earlier terms found becomes

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

For $n=4$ the recurrence equation gives

$$a_4(4+r)+a_3=0$$

Which after substituting the earlier terms found becomes

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

For $n=5$ the recurrence equation gives

$$a_5(5+r)+a_4=0$$

Which after substituting the earlier terms found becomes

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

And so on. Therefore the solution is

$$\begin{array}{rl}y& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r}\\ & =a_0{x}^r+a_1{x}^{1+r}+a_2{x}^{2+r}+a_3{x}^{3+r}+\dots \end{array}$$

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

$$y=a_0{x}^r-\frac{a_0{x}^{1+r}}{1+r}+\frac{a_0{x}^{2+r}}{(1+r)(2+r)}-\frac{a_0{x}^{3+r}}{(1+r)(2+r)(3+r)}+\frac{a_0{x}^{4+r}}{(1+r)(2+r)(3+r)(4+r)}-\frac{a_0{x}^{5+r}}{(1+r)(2+r)(3+r)(4+r)(5+r)}+\dots$$

Which can be written as

$$y=x^r(a_0-\frac{a_{0}x}{1+r}+\frac{a_0{x}^2}{(1+r)(2+r)}-\frac{a_0{x}^3}{(1+r)(2+r)(3+r)}+\frac{a_0{x}^4}{(1+r)(2+r)(3+r)(4+r)}-\frac{a_0{x}^5}{(1+r)(2+r)(3+r)(4+r)(5+r)}+O\left(x^6\right)a_0)$$

Collecting terms, the solution becomes

$$\begin{array}{}\text{(3)}& y=\left[x^r(1-\frac{x}{1+r}+\frac{x^2}{(1+r)(2+r)}-\frac{x^3}{(1+r)(2+r)(3+r)}+\frac{x^4}{(1+r)(2+r)(3+r)(4+r)}-\frac{x^5}{(1+r)(2+r)(3+r)(4+r)(5+r)}+O\left(x^6\right))a_0\right]\end{array}$$

Finally, since $r=0$, then the solution becomes

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

Unable to solve the balance equation $m{c}_0{x}^{-1+m}=\frac{1}{x}$ for $c_0$ and $x$. No particular solution exists.

Unable to find the particular solution. No solution exist.

✗ Maple

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

$$\text{No solution found}$$

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}y\left(x\right)+y\left(x\right)=\frac{1}{x}\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=-y\left(x\right)+\frac{1}{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}\\ & & \frac{d}{dx}y\left(x\right)+y\left(x\right)=\frac{1}{x}\\ \bullet & & \text{The ODE is linear; multiply by an integrating factor}\mu \left(x\right)\\ & & \mu \left(x\right)(\frac{d}{dx}y\left(x\right)+y\left(x\right))=\frac{\mu \left(x\right)}{x}\\ \bullet & & \text{Assume the lhs of the ODE is the total derivative}\frac{d}{dx}\left(y\left(x\right)\mu \left(x\right)\right)\\ & & \mu \left(x\right)(\frac{d}{dx}y\left(x\right)+y\left(x\right))=\left(\frac{d}{dx}y\left(x\right)\right)\mu \left(x\right)+y\left(x\right)\left(\frac{d}{dx}\mu \left(x\right)\right)\\ \bullet & & \text{Isolate}\frac{d}{dx}\mu \left(x\right)\\ & & \frac{d}{dx}\mu \left(x\right)=\mu \left(x\right)\\ \bullet & & \text{Solve to find the integrating factor}\\ & & \mu \left(x\right)={\mathrm{e}}^x\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \left(\frac{d}{dx}\left(y\left(x\right)\mu \left(x\right)\right)\right)dx=\int \frac{\mu \left(x\right)}{x}dx+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate the integral on the lhs}\\ & & y\left(x\right)\mu \left(x\right)=\int \frac{\mu \left(x\right)}{x}dx+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)=\frac{\int \frac{\mu \left(x\right)}{x}dx+\mathit{C}\mathit{1}}{\mu \left(x\right)}\\ \bullet & & \text{Substitute}\mu \left(x\right)={\mathrm{e}}^x\\ & & y\left(x\right)=\frac{\int \frac{{\mathrm{e}}^x}{x}dx+\mathit{C}\mathit{1}}{{\mathrm{e}}^x}\\ \bullet & & \text{Evaluate the integrals on the rhs}\\ & & y\left(x\right)=\frac{-{\mathrm{Ei}}_1(-x)+\mathit{C}\mathit{1}}{{\mathrm{e}}^x}\\ \bullet & & \text{Simplify}\\ & & y\left(x\right)={\mathrm{e}}^{-x}(-{\mathrm{Ei}}_1(-x)+\mathit{C}\mathit{1})\end{array}$$

✓ Mathematica. Time used: 0.014 (sec). Leaf size: 113

ode=D[y[x],x]+y[x]==1/x; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

$$y(x)\to (-\frac{x^5}{120}+\frac{x^4}{24}-\frac{x^3}{6}+\frac{x^2}{2}-x+1)(\frac{x^6}{2160}+\frac{x^5}{600}+\frac{x^4}{96}+\frac{x^3}{18}+\frac{x^2}{4}+x+\log (x))+c_1(-\frac{x^5}{120}+\frac{x^4}{24}-\frac{x^3}{6}+\frac{x^2}{2}-x+1)$$

✗ Sympy

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

ValueError : ODE y(x) + Derivative(y(x), x) - 1/x does not match hint 1st_power_series