2.5.9 problem 9

Internal problem ID [8646]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 9
Date solved : Tuesday, December 17, 2024 at 12:56:48 PM
CAS classification : [_separable]

Solve

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

Using series expansion around \(x=0\)

Let the solution be represented as Frobenius power series of the form

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

Then

\begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \end{align*}

Substituting the above back into the ode gives

\begin{align*} \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\tag {1} \end{align*}

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

\begin{align*} \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 \end{align*}

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} x^r \end{align*}

At \(x = 0\) the solution above becomes

\begin{gather*} y = c_1 \left (\frac {1}{x}+O\left (x^{6}\right )\right ) \end{gather*}

Figure 2.224: Slope field plot
\(x y^{\prime }+y = 0\)

Maple step by step solution

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

Maple trace

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

Maple dsolve solution

Solving time : 0.018 (sec)
Leaf size : 14

dsolve(diff(y(x),x)*x+y(x) = 0,y(x), 
       series,x=0)
 

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

Mathematica DSolve solution

Solving time : 0.002 (sec)
Leaf size : 9

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

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