2.4.10 Problem 10

Internal problem ID [10355]
Book : First order enumerated odes
Section : section 4. First order odes solved using series method
Problem number : 10
Date solved : Monday, March 09, 2026 at 03:35:21 AM
CAS classification : [_separable]

\begin{align*} y^{\prime }+\frac {y}{x}&=0 \\ \end{align*}

Series expansion around \(x=0\).

Entering first order ode series solverEntering first order ode series solver frobenius solverLet the homogeneous solution be represented as Frobenius power series of the form

\[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \]

Then

\[ y^{\prime } = \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \]

Substituting the above back into the ode gives

\begin{equation} \tag{1} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}}{x} = 0 \end{equation}

Which simplifies to

\begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n}\right ) = 0 \end{equation}

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 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{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n}\right ) = 0 \end{equation}

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

\[ \left (n +r \right ) a_{n} x^{n +r -1}+x^{n +r -1} a_{n} = 0 \]

When \(n=0\) the above becomes

\[ r a_{0} x^{-1+r}+x^{-1+r} a_{0} = 0 \]

Since \(a_{0}\neq 0\) then the indicial equation becomes

\[ \left (r +1\right ) x^{-1+r} = 0 \]

Since the above is true for all \(x\) then the indicial equation simplifies to

\[ r +1 = 0 \]

Solving for \(r\) gives the root of the indicial equation as

\[ r=-1 \]

Replacing \(r=-1\) found above results in

\[ \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} a_{n}\right ) = 0 \]

From the above we see that there is no recurrence relation since there is only one summation term. Therefore all \(a_{n}\) terms are zero except for \(a_{0}\). Hence

\begin{align*} y_h &= a_{0} \left (\frac {1}{x}+O\left (x^{6}\right )\right ) \end{align*}

The solution is

\[ y = c_1 \left (\frac {1}{x}+O\left (x^{6}\right )\right ) \]

Direction field \(y^{\prime }+\frac {y}{x} = 0\)	Isoclines for \(y^{\prime }+\frac {y}{x} = 0\)

2.4.10.1 ✓ Maple. Time used: 0.004 (sec). Leaf size: 14

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

\[ y = \frac {c_1}{x}+O\left (x^{6}\right ) \]

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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )+\frac {y \left (x \right )}{x}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right )}{x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}=-\frac {1}{x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}d x =\int -\frac {1}{x}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y \left (x \right )\right )=-\ln \left (x \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {{\mathrm e}^{\mathit {C1}}}{x} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\mathit {C1}}{x} \end {array} \]

2.4.10.2 ✓ Mathematica. Time used: 0.001 (sec). Leaf size: 9

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

\[ y(x)\to \frac {c_1}{x} \]

2.4.10.3 ✗ Sympy

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

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

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('separable', '1st_exact', '1st_linear', 'Bernoulli', '1st_homogeneous_coeff_best', '1st_homogeneous_coeff_subs_indep_div_dep', '1st_homogeneous_coeff_subs_dep_div_indep', 'almost_linear', 'lie_group', 'nth_linear_euler_eq_homogeneous', 'separable_Integral', '1st_exact_Integral', '1st_linear_Integral', 'Bernoulli_Integral', '1st_homogeneous_coeff_subs_indep_div_dep_Integral', '1st_homogeneous_coeff_subs_dep_div_indep_Integral', 'almost_linear_Integral')