2.5.9 Problem 9

Internal problem ID [8970]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 9
Date solved : Sunday, March 30, 2025 at 01:57:35 PM
CAS classification : [_separable]

Solve

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

Using series expansion around $x=0$

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

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

Which simplifies to

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

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

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

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

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

When $n=0$ the above becomes

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

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

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

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

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

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

$$\begin{array}{rl}{y}_h& =a_0{x}^r\end{array}$$

At $x=0$ the solution above becomes

$$\begin{array}{c}y=c_1(\frac{1}{x}+O\left(x^6\right))\end{array}$$

Figure 2.193: Slope field $x{y}^{\prime }+y=0$

✓ Maple. Time used: 0.022 (sec). Leaf size: 14

Order:=6; 
ode:=diff(y(x),x)*x+y(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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x\left(\frac{d}{dx}y\left(x\right)\right)+y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dx}y\left(x\right)}{y\left(x\right)}=-\frac{1}{x}\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}y\left(x\right)}{y\left(x\right)}dx=\int -\frac{1}{x}dx+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & \ln \left(y\left(x\right)\right)=-\ln \left(x\right)+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)=\frac{{\mathrm{e}}^{\mathit{C}\mathit{1}}}{x}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(x\right)=\frac{\mathit{C}\mathit{1}}{x}\end{array}$$

✓ Mathematica. Time used: 0.002 (sec). Leaf size: 9

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

$$y(x)\to \frac{c_1}{x}$$

✗ Sympy

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

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