17.15 problem 2(b) solving using series

Internal problem ID [6418]
Internal file name [OUTPUT/5666_Sunday_June_05_2022_03_46_26_PM_70853944/index.tex]

Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Section 4.2. Series Solutions of First-Order Differential Equations Page 162
Problem number: 2(b) solving using series.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "first order ode series method. Irregular singular point"

Maple gives the following as the ode type

[_separable]

Unable to solve or complete the solution.

\[ \boxed {x^{2} y^{\prime }-y=0} \] With the expansion point for the power series method at \(x = 0\).

17.15.1 Solving as series ode

Writing the ODE as \begin {align*} y^{\prime } + q(x)y &= p(x) \\ y^{\prime }-\frac {y}{x^{2}} &= 0 \end {align*}

Where \begin {align*} q(x) &= -\frac {1}{x^{2}}\\ p(x) &= 0 \end {align*}

Next, the type of the expansion point \(x = 0\) is determined. This point can be an ordinary point, a regular singular point (also called removable singularity), or irregular singular point (also called non-removable singularity or essential singularity). When \(x = 0\) is an ordinary point, then the standard power series is used. If the point is a regular singular point, Frobenius series is used instead. Irregular singular point requires more advanced methods (asymptotic methods) and is not supported now. Hopefully this will be added in the future. \(x = 0\) is called an ordinary point \(q(x)\) has a Taylor series expansion around the point \(x = 0\). \(x = 0\) is called a regular singular point if \(q(x)\) is not not analytic at \(x = 0\) but \(x q(x)\) has Taylor series expansion. And finally, \(x = 0\) is an irregular singular point if the point is not ordinary and not regular singular. This is the most complicated case. Now the expansion point \(x = 0\) is checked to see if it is an ordinary point or not.

Since \(x = 0\) is not an ordinary point, we now check to see if it is a regular singular point. \[ xq(x)=-\frac {1}{x} \] does not have a Taylor series around \(x = 0\).

Unable to solve since \(x = 0\) is not regular singular point. Terminating.

Verification of solutions N/A

17.15.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{x^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {1}{x^{2}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {1}{x^{2}}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=c_{1} -\frac {1}{x} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{\frac {c_{1} x -1}{x}} \end {array} \]

Maple trace

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

✗ Solution by Maple

Order:=8; 
dsolve(x^2*diff(y(x),x)=y(x),y(x),type='series',x=0);
 

\[ \text {No solution found} \]

✓ Solution by Mathematica

Time used: 0.008 (sec). Leaf size: 13

AsymptoticDSolveValue[x^2*y'[x]==y[x],y[x],{x,0,7}]
 

\[ y(x)\to c_1 e^{-1/x} \]