2.4.59 Problem 56

Internal problem ID [8947]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 56
Date solved : Friday, April 25, 2025 at 05:25:31 PM
CAS classification : [[_Emden, _Fowler]]

Solve

$$\begin{array}{rl}2x^2{y}^{\prime \prime }+5x{y}^{\prime }+4y& =0\end{array}$$

Using series expansion around $x=0$

The type of the expansion point is first determined. This is done on the homogeneous part of the ODE.

$$2x^2{y}^{\prime \prime }+5x{y}^{\prime }+4y=0$$

The following is summary of singularities for the above ode. Writing the ode as

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

Where

$$\begin{array}{rl}p(x)& =\frac{5}{2x}\\ q(x)& =\frac{2}{x^2}\end{array}$$

Table 2.110: Table $p(x),q(x)$ singularites.

$p(x)=\frac{5}{2x}$
singularity	type
$x=0$	$\text{“regular”}$

$q(x)=\frac{2}{x^2}$
singularity	type
$x=0$	$\text{“regular”}$

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : $[0,\mathrm{\infty }]$

Irregular singular points : $[]$

Since $x=0$ is regular singular point, then Frobenius power series is used. The ode is normalized to be

$$2x^2{y}^{\prime \prime }+5x{y}^{\prime }+4y=0$$

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

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

Substituting the above back into the ode gives

$$\begin{array}{}\text{(1)}& 2x^2(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2})+5x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)a_n{x}^{n+r-1})+4(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r})=0\end{array}$$

Which simplifies to

$$\begin{array}{}\text{(2A)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}2x^{n+r}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}5x^{n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}4a_n{x}^{n+r})=0\end{array}$$

The next step is to make all powers of $x$ be $n+r$ in each summation term. Going over each summation term above with power of $x$ in it which is not already $x^{n+r}$ 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 $n+r$.

$$\begin{array}{}\text{(2B)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}2x^{n+r}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}5x^{n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}4a_n{x}^{n+r})=0\end{array}$$

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

$$2x^{n+r}{a}_n(n+r)(n+r-1)+5x^{n+r}{a}_n(n+r)+4a_n{x}^{n+r}=0$$

When $n=0$ the above becomes

$$2x^r{a}_{0}r(-1+r)+5x^r{a}_{0}r+4a_0{x}^r=0$$

Or

$$(2x^{r}r(-1+r)+5x^{r}r+4x^r)a_0=0$$

Since $a_0\ne 0$ then the above simplifies to

$$(2r^2+3r+4)x^r=0$$

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

$$2r^2+3r+4=0$$

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

$$\begin{array}{rl}{r}_1& =-\frac{3}{4}+\frac{i\sqrt{23}}{4}\\ r_2& =-\frac{3}{4}-\frac{i\sqrt{23}}{4}\end{array}$$

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

$$(2r^2+3r+4)x^r=0$$

Solving for $r$ gives the roots of the indicial equation as $[-\frac{3}{4}+\frac{i\sqrt{23}}{4},-\frac{3}{4}-\frac{i\sqrt{23}}{4}]$.

Since the roots are complex conjugates, then two linearly independent solutions can be constructed using

$$\begin{array}{rl}{y}_1\left(x\right)& =x^{r_1}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^n)\\ y_2\left(x\right)& =x^{r_2}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)\end{array}$$

Or

$$\begin{array}{rl}{y}_1\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n-\frac{3}{4}+\frac{i\sqrt{23}}{4}}\\ y_2\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n-\frac{3}{4}-\frac{i\sqrt{23}}{4}}\end{array}$$

$y_1\left(x\right)$ is found first. Eq (2B) derived above is now used to find all $a_n$ coefficients. The case $n=0$ is skipped since it was used to find the roots of the indicial equation. $a_0$ is arbitrary and taken as $a_0=1$. For $0\le n$ the recursive equation is

$$\begin{array}{}\text{(3)}& 2a_n(n+r)(n+r-1)+5a_n(n+r)+4a_n=0\end{array}$$

Solving for $a_n$ from recursive equation (4) gives

$$\begin{array}{}\text{(4)}& a_n=0\end{array}$$

Which for the root $r=-\frac{3}{4}+\frac{i\sqrt{23}}{4}$ becomes

$$\begin{array}{}\text{(5)}& a_n=0\end{array}$$

At this point, it is a good idea to keep track of $a_n$ in a table both before substituting $r=-\frac{3}{4}+\frac{i\sqrt{23}}{4}$ and after as more terms are found using the above recursive equation.

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$

For $n=1$, using the above recursive equation gives

$$a_1=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$

For $n=2$, using the above recursive equation gives

$$a_2=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$0$	$0$

For $n=3$, using the above recursive equation gives

$$a_3=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$0$	$0$
$a_3$	$0$	$0$

For $n=4$, using the above recursive equation gives

$$a_4=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$0$	$0$
$a_3$	$0$	$0$
$a_4$	$0$	$0$

For $n=5$, using the above recursive equation gives

$$a_5=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$0$	$0$
$a_3$	$0$	$0$
$a_4$	$0$	$0$
$a_5$	$0$	$0$

Using the above table, then the solution $y_1\left(x\right)$ is

$$\begin{array}{rl}{y}_1\left(x\right)& =x^{-\frac{3}{4}+\frac{i\sqrt{23}}{4}}(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots )\\ & =x^{-\frac{3}{4}+\frac{i\sqrt{23}}{4}}(1+O\left(x^6\right))\end{array}$$

The second solution $y_2\left(x\right)$ is found by taking the complex conjugate of $y_1\left(x\right)$ which gives

$$y_2\left(x\right)=x^{-\frac{3}{4}-\frac{i\sqrt{23}}{4}}(1+O\left(x^6\right))$$

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =c_1{x}^{-\frac{3}{4}+\frac{i\sqrt{23}}{4}}(1+O\left(x^6\right))+c_2{x}^{-\frac{3}{4}-\frac{i\sqrt{23}}{4}}(1+O\left(x^6\right))\end{array}$$

Hence the final solution is

$$\begin{array}{rl}y& =y_h\\ & =c_1{x}^{-\frac{3}{4}+\frac{i\sqrt{23}}{4}}(1+O\left(x^6\right))+c_2{x}^{-\frac{3}{4}-\frac{i\sqrt{23}}{4}}(1+O\left(x^6\right))\end{array}$$

✓ Maple. Time used: 0.031 (sec). Leaf size: 32

Order:=6; 
ode:=2*x^2*diff(diff(y(x),x),x)+5*x*diff(y(x),x)+4*y(x) = 0; 
dsolve(ode,y(x),type='series',x=0);
 

$$y=\frac{x^{-\frac{i\sqrt{23}}{4}}{c}_1+x^{\frac{i\sqrt{23}}{4}}{c}_2}{x^{3/4}}+O\left(x^6\right)$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
<- LODE of Euler type successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 2x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+5x\left(\frac{d}{dx}y\left(x\right)\right)+4y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=-\frac{2y\left(x\right)}{x^2}-\frac{5\left(\frac{d}{dx}y\left(x\right)\right)}{2x}\\ \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; ODE is linear}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)+\frac{5\left(\frac{d}{dx}y\left(x\right)\right)}{2x}+\frac{2y\left(x\right)}{x^2}=0\\ \bullet & & \text{Multiply by denominators of the ODE}\\ & & 2x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+5x\left(\frac{d}{dx}y\left(x\right)\right)+4y\left(x\right)=0\\ \bullet & & \text{Make a change of variables}\\ & & t=\ln \left(x\right)\\ ◻& & \text{Substitute the change of variables back into the ODE}\\ & \circ & \text{Calculate the}\text{1st}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{d}{dx}y\left(x\right)=\left(\frac{d}{dt}y\left(t\right)\right)\left(\frac{d}{dx}t\left(x\right)\right)\\ & \circ & \text{Compute derivative}\\ & & \frac{d}{dx}y\left(x\right)=\frac{\frac{d}{dt}y\left(t\right)}{x}\\ & \circ & \text{Calculate the}\text{2nd}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right){\left(\frac{d}{dx}t\left(x\right)\right)}^2+\left(\frac{d}{dx}\frac{d}{dx}t\left(x\right)\right)\left(\frac{d}{dt}y\left(t\right)\right)\\ & \circ & \text{Compute derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\frac{\frac{d}{dt}\frac{d}{dt}y\left(t\right)}{x^2}-\frac{\frac{d}{dt}y\left(t\right)}{x^2}\\ & & \text{Substitute the change of variables back into the ODE}\\ & & 2x^2(\frac{\frac{d}{dt}\frac{d}{dt}y\left(t\right)}{x^2}-\frac{\frac{d}{dt}y\left(t\right)}{x^2})+5\frac{d}{dt}y\left(t\right)+4y\left(t\right)=0\\ \bullet & & \text{Simplify}\\ & & 2\frac{d}{dt}\frac{d}{dt}y\left(t\right)+3\frac{d}{dt}y\left(t\right)+4y\left(t\right)=0\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)=-\frac{3\frac{d}{dt}y\left(t\right)}{2}-2y\left(t\right)\\ \bullet & & \text{Group terms with}y\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)+\frac{3\frac{d}{dt}y\left(t\right)}{2}+2y\left(t\right)=0\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2+\frac{3}{2}r+2=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{(-\frac{3}{2})\pm \left(\sqrt{-\frac{23}{4}}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-\frac{3}{4}-\frac{\mathrm{I}\sqrt{23}}{4},-\frac{3}{4}+\frac{\mathrm{I}\sqrt{23}}{4})\\ \bullet & & \text{1st solution of the ODE}\\ & & y_1\left(t\right)={\mathrm{e}}^{-\frac{3t}{4}}\cos \left(\frac{\sqrt{23}t}{4}\right)\\ \bullet & & \text{2nd solution of the ODE}\\ & & y_2\left(t\right)={\mathrm{e}}^{-\frac{3t}{4}}\sin \left(\frac{\sqrt{23}t}{4}\right)\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{y}_1\left(t\right)+\mathit{C}\mathit{2}{y}_2\left(t\right)\\ \bullet & & \text{Substitute in solutions}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-\frac{3t}{4}}\cos \left(\frac{\sqrt{23}t}{4}\right)+\mathit{C}\mathit{2}{\mathrm{e}}^{-\frac{3t}{4}}\sin \left(\frac{\sqrt{23}t}{4}\right)\\ \bullet & & \text{Change variables back using}t=\ln \left(x\right)\\ & & y\left(x\right)=\frac{\mathit{C}\mathit{1}\cos \left(\frac{\sqrt{23}\ln \left(x\right)}{4}\right)}{x^{3/4}}+\frac{\mathit{C}\mathit{2}\sin \left(\frac{\sqrt{23}\ln \left(x\right)}{4}\right)}{x^{3/4}}\\ \bullet & & \text{Simplify}\\ & & y\left(x\right)=\frac{\mathit{C}\mathit{1}\cos \left(\frac{\sqrt{23}\ln \left(x\right)}{4}\right)}{x^{3/4}}+\frac{\mathit{C}\mathit{2}\sin \left(\frac{\sqrt{23}\ln \left(x\right)}{4}\right)}{x^{3/4}}\end{array}$$

✓ Mathematica. Time used: 0.005 (sec). Leaf size: 44

ode=2*x^2*D[y[x],{x,2}]+5*x*D[y[x],x]+4*y[x]==0; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

$$y(x)\to c_1{x}^{\frac{1}{4}(-3+i\sqrt{23})}+c_2{x}^{\frac{1}{4}(-3-i\sqrt{23})}$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x**2*Derivative(y(x), (x, 2)) + 5*x*Derivative(y(x), x) + 4*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
 

ValueError : Expected Expr or iterable but got None