2.2.16 Problem 17

Internal problem ID [9138]
Book : Second order enumerated odes
Section : section 2
Problem number : 17
Date solved : Friday, April 25, 2025 at 05:57:14 PM
CAS classification : [[_Emden, _Fowler], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

Solved as second order ode using change of variable on x method 2

Time used: 0.254 (sec)

Solve

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

In normal form the ode

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+\frac{2y^{\prime }}{x}+\frac{a^{2}y}{x^4}& =0\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =\frac{2}{x}\\ q\left(x\right)& =\frac{a^2}{x^4}\end{array}$$

Applying change of variables $\tau =g\left(x\right)$ to (2) gives

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

Let $p_1=0$. Eq (4) simplifies to

$$\begin{array}{rl}{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)& =0\end{array}$$

This ode is solved resulting in

$$\begin{array}{rl}\tau & =\int {\mathrm{e}}^{-\int p\left(x\right)dx}dx\\ & =\int {\mathrm{e}}^{-\int \frac{2}{x}dx}dx\\ & =\int e^{-2\ln \left(x\right)}dx\\ & =\int \frac{1}{x^2}dx\\ \text{(6)}& & =-\frac{1}{x}\end{array}$$

Using (6) to evaluate $q_1$ from (5) gives

$$\begin{array}{rl}{q}_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{\frac{a^2}{x^4}}{\frac{1}{x^4}}\\ \text{(7)}& & =a^2\end{array}$$

Substituting the above in (3) and noting that now $p_1=0$ results in

$$\begin{array}{rl}\frac{d^2}{d{\tau }^2}y\left(\tau \right)+q_{1}y\left(\tau \right)& =0\\ \frac{d^2}{d{\tau }^2}y\left(\tau \right)+a^{2}y\left(\tau \right)& =0\end{array}$$

The above ode is now solved for $y\left(\tau \right)$.This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(\tau )+B{y}^{\prime }(\tau )+Cy(\tau )=0$$

Where in the above $A=1,B=0,C=a^2$. Let the solution be $y=e^{\lambda \tau }$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{\tau \lambda }+a^2{\mathrm{e}}^{\tau \lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda \tau }$ gives

$$\begin{array}{}\text{(2)}& a^2+{\lambda }^2=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=0,C=a^2$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{0}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{0^2-(4)\left(1\right)\left(a^2\right)}\\ & =\pm \sqrt{-a^2}\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =+\sqrt{-a^2}\\ {\lambda }_2& =-\sqrt{-a^2}\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =ia\\ {\lambda }_2& =-ia\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =0$ and $\beta =a$. Therefore the final solution, when using Euler relation, can be written as

$$y=e^{\alpha \tau }(c_1\cos (\beta \tau )+c_2\sin (\beta \tau ))$$

Which becomes

$$y=e^0(c_1\cos \left(a\tau \right)+c_2\sin \left(a\tau \right))$$

Or

$$y=c_1\cos \left(a\tau \right)+c_2\sin \left(a\tau \right)$$

Will add steps showing solving for IC soon.

The above solution is now transformed back to $y\left(x\right)$ using (6) which results in

$$y\left(x\right)=c_1\cos \left(\frac{a}{x}\right)-c_2\sin \left(\frac{a}{x}\right)$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y\left(x\right)& =c_1\cos \left(\frac{a}{x}\right)-c_2\sin \left(\frac{a}{x}\right)\end{array}$$

Solved as second order ode using change of variable on x method 1

Time used: 0.201 (sec)

Solve

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

In normal form the ode

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+\frac{2y^{\prime }}{x}+\frac{a^{2}y}{x^4}& =0\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =\frac{2}{x}\\ q\left(x\right)& =\frac{a^2}{x^4}\end{array}$$

Applying change of variables $\tau =g\left(x\right)$ to (2) results

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

Let $q_1=c^2$ where $c$ is some constant. Therefore from (5)

$$\begin{array}{rl}{\tau }^{\prime }& =\frac{1}{c}\sqrt{q}\\ \text{(6)}& & =\frac{\sqrt{\frac{a^2}{x^4}}}{c}{\tau }^{″}& =-\frac{2a^2}{c\sqrt{\frac{a^2}{x^4}}{x}^5}\end{array}$$

Substituting the above into (4) results in

$$\begin{array}{rl}{p}_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{-\frac{2a^2}{c\sqrt{\frac{a^2}{x^4}}{x}^5}+\frac{2}{x}\frac{\sqrt{\frac{a^2}{x^4}}}{c}}{{\left(\frac{\sqrt{\frac{a^2}{x^4}}}{c}\right)}^2}\\ & =0\end{array}$$

Therefore ode (3) now becomes

$$\begin{array}{rl}y{\left(\tau \right)}^{″}+p_{1}y{\left(\tau \right)}^{\prime }+q_{1}y\left(\tau \right)& =0\\ \text{(7)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+c^{2}y\left(\tau \right)& =0\end{array}$$

The above ode is now solved for $y\left(\tau \right)$. Since the ode is now constant coefficients, it can be easily solved to give

$$\begin{array}{rl}y\left(\tau \right)& =c_1\cos \left(c\tau \right)+c_2\sin \left(c\tau \right)\end{array}$$

Now from (6)

$$\begin{array}{rl}\tau & =\int \frac{1}{c}\sqrt{q}dx\\ & =\frac{\int \sqrt{\frac{a^2}{x^4}}dx}{c}\\ & =-\frac{x\sqrt{\frac{a^2}{x^4}}}{c}\end{array}$$

Substituting the above into the solution obtained gives

$$y=c_1\cos \left(x\sqrt{\frac{a^2}{x^4}}\right)-c_2\sin \left(x\sqrt{\frac{a^2}{x^4}}\right)$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\cos \left(x\sqrt{\frac{a^2}{x^4}}\right)-c_2\sin \left(x\sqrt{\frac{a^2}{x^4}}\right)\end{array}$$

Solved as second order Bessel ode

Time used: 0.101 (sec)

Solve

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

Writing the ode as

$$\begin{array}{}\text{(1)}& x^2{y}^{\prime \prime }+2y^{\prime }x+\frac{a^{2}y}{x^2}=0\end{array}$$

Bessel ode has the form

$$\begin{array}{}\text{(2)}& x^2{y}^{\prime \prime }+y^{\prime }x+(-n^2+x^2)y=0\end{array}$$

The generalized form of Bessel ode is given by Bowman (1958) as the following

$$\begin{array}{}\text{(3)}& x^2{y}^{\prime \prime }+(1-2\alpha )x{y}^{\prime }+({\beta }^2{\gamma }^2{x}^{2\gamma }-n^2{\gamma }^2+{\alpha }^2)y=0\end{array}$$

With the standard solution

$$\begin{array}{}\text{(4)}& y& =x^{\alpha }(c_1\mathrm{BesselJ}(n,\beta x^{\gamma })+c_2\mathrm{BesselY}(n,\beta x^{\gamma }))\end{array}$$

Comparing (3) to (1) and solving for $\alpha ,\beta ,n,\gamma$ gives

$$\begin{array}{rl}\alpha & =-\frac{1}{2}\\ \beta & =a\\ n& =\frac{1}{2}\\ \gamma & =-1\end{array}$$

Substituting all the above into (4) gives the solution as

$$\begin{array}{r}y=\frac{c_1\sqrt{2}\sin \left(\frac{a}{x}\right)}{\sqrt{x}\sqrt{\pi }\sqrt{\frac{a}{x}}}-\frac{c_2\sqrt{2}\cos \left(\frac{a}{x}\right)}{\sqrt{x}\sqrt{\pi }\sqrt{\frac{a}{x}}}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{c_1\sqrt{2}\sin \left(\frac{a}{x}\right)}{\sqrt{x}\sqrt{\pi }\sqrt{\frac{a}{x}}}-\frac{c_2\sqrt{2}\cos \left(\frac{a}{x}\right)}{\sqrt{x}\sqrt{\pi }\sqrt{\frac{a}{x}}}\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.298 (sec)

Solve

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

Writing the ode as

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+\frac{2y^{\prime }}{x}+\frac{a^{2}y}{x^4}& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =\frac{2}{x}C& =\frac{a^2}{x^4}\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =y{e}^{\int \frac{B}{2A}dx}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(x)=rz(x)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{-a^2}{x^4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-a^2\\ t& =x^4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(-\frac{a^2}{x^4})z(x)\end{array}$$

Equation (7) is now solved. After finding $z(x)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.33: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-0\\ & =4\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=x^4$. There is a pole at $x=0$ of order $4$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $4$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Attempting to find a solution using case $n=1$.

Looking at higher order poles of order $2v$≥$4$ (must be even order for case one).Then for each pole $c$, $[\sqrt{r}{]}_c$ is the sum of terms $\frac{1}{(x-c{)}^i}$ for $2\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ expanded around each pole $c$. Hence

$$\begin{array}{}\text{(1B)}& [\sqrt{r}{]}_c& =\sum _2^v\frac{a_i}{(x-c{)}^i}\end{array}$$

Let $a$ be the coefficient of the term $\frac{1}{(x-c{)}^v}$ in the above where $v$ is the pole order divided by 2. Let $b$ be the coefficient of $\frac{1}{(x-c{)}^{v+1}}$ in $r$ minus the coefficient of $\frac{1}{(x-c{)}^{v+1}}$ in $[\sqrt{r}{]}_c$. Then

$$\begin{array}{rl}{\alpha }_c^{+}& =\frac{1}{2}(\frac{b}{a}+v)\\ {\alpha }_c^{-}& =\frac{1}{2}(-\frac{b}{a}+v)\end{array}$$

The partial fraction decomposition of $r$ is

$$r=-\frac{a^2}{x^4}$$

There is pole in $r$ at $x=0$ of order $4$, hence $v=2$. Expanding $\sqrt{r}$ as Laurent series about this pole $c=0$ gives

$$\begin{array}{}\text{(2B)}& [\sqrt{r}{]}_c\approx \frac{ia}{x^2}+\dots \end{array}$$

Using eq. (1B), taking the sum up to $v=2$ the above becomes

$$\begin{array}{}\text{(3B)}& [\sqrt{r}{]}_c=\frac{ia}{x^2}\end{array}$$

The above shows that the coefficient of $\frac{1}{(x-0{)}^2}$ is

$$a=ia$$

Now we need to find $b$. let $b$ be the coefficient of the term $\frac{1}{(x-c{)}^{v+1}}$ in $r$ minus the coefficient of the same term but in the sum $[\sqrt{r}{]}_c$ found in eq. (3B). Here $c$ is current pole which is $c=0$. This term becomes $\frac{1}{x^3}$. The coefficient of this term in the sum $[\sqrt{r}{]}_c$ is seen to be $0$ and the coefficient of this term $r$ is found from the partial fraction decomposition from above to be $0$. Therefore

$$\begin{array}{rl}b& =\left(0\right)-(0)\\ & =0\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =\frac{ia}{x^2}\\ {\alpha }_c^{+}& =\frac{1}{2}(\frac{b}{a}+v)& & =\frac{1}{2}(\frac{0}{ia}+2)& & =1\\ {\alpha }_c^{-}& =\frac{1}{2}(-\frac{b}{a}+v)& & =\frac{1}{2}(-\frac{0}{ia}+2)& & =1\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is $4>2$ then

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =0\\ {\alpha }_{\mathrm{\infty }}^{-}& =1\end{array}$$

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ where $r$ is

$$r=-\frac{a^2}{x^4}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$4$	$\frac{ia}{x^2}$	$1$	$1$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$4$	$0$	$0$	$1$

Now that the all $[\sqrt{r}{]}_c$ and its associated ${\alpha }_c^{\pm }$ have been determined for all the poles in the set $\mathrm{\Gamma }$ and $[\sqrt{r}{]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$ have also been found, the next step is to determine possible non negative integer $d$ from these using

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{s(\mathrm{\infty })}-\sum _{c\in \mathrm{\Gamma }}{\alpha }_c^{s(c)}\end{array}$$

Where $s(c)$ is either $+$ or $-$ and $s(\mathrm{\infty })$ is the sign of ${\alpha }_{\mathrm{\infty }}^{\pm }$. This is done by trial over all set of families $s=(s(c){)}_{c\in \mathrm{\Gamma }\cup \mathrm{\infty }}$ until such $d$ is found to work in finding candidate $\omega$. Trying ${\alpha }_{\mathrm{\infty }}^{-}=1$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)\\ & =1-\left(1\right)\\ & =0\end{array}$$

Since $d$ an integer and $d\ge 0$ then it can be used to find $\omega$ using

$$\begin{array}{rl}\omega & =\sum _{c\in \mathrm{\Gamma }}(s(c)[\sqrt{r}{]}_c+\frac{{\alpha }_c^{s(c)}}{x-c})+s(\mathrm{\infty })[\sqrt{r}{]}_{\mathrm{\infty }}\end{array}$$

The above gives

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{ia}{x^2}+\frac{1}{x}+(-)\left(0\right)\\ & =-\frac{ia}{x^2}+\frac{1}{x}\\ & =\frac{-ia+x}{x^2}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=0$ to solve the ode. The polynomial $p(x)$ needs to satisfy the equation

$$\begin{array}{}\text{(1A)}& p^{″}+2\omega p^{\prime }+({\omega }^{\prime }+{\omega }^2-r)p=0\end{array}$$

Let

$$\begin{array}{}\text{(2A)}& p(x)& =1\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{ia}{x^2}+\frac{1}{x})\left(0\right)+((\frac{2ia}{x^3}-\frac{1}{x^2})+{(-\frac{ia}{x^2}+\frac{1}{x})}^2-(-\frac{a^2}{x^4}))& =0\\ 0=0\end{array}$$

The equation is satisfied since both sides are zero. Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (-\frac{ia}{x^2}+\frac{1}{x})dx}\\ & =x{\mathrm{e}}^{\frac{ia}{x}}\end{array}$$

The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{\frac{2}{x}}{1}dx}\\ & =z_1{e}^{-\ln \left(x\right)}\\ & =z_1\left(\frac{1}{x}\right)\end{array}$$

Which simplifies to

$$y_1={\mathrm{e}}^{\frac{ia}{x}}$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx$$

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{\frac{2}{x}}{1}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-2\ln \left(x\right)}}{{\left(y_1\right)}^2}dx\\ & =y_1(-\frac{i{\mathrm{e}}^{-\frac{2ia}{x}}}{2a})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left({\mathrm{e}}^{\frac{ia}{x}}\right)+c_2\left({\mathrm{e}}^{\frac{ia}{x}}(-\frac{i{\mathrm{e}}^{-\frac{2ia}{x}}}{2a})\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1{\mathrm{e}}^{\frac{ia}{x}}-\frac{i{c}_2{\mathrm{e}}^{-\frac{ia}{x}}}{2a}\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 21

ode:=diff(diff(y(x),x),x)+2/x*diff(y(x),x)+a^2/x^4*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1\sin \left(\frac{a}{x}\right)+c_2\cos \left(\frac{a}{x}\right)$$

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 
trying a symmetry of the form [xi=0, eta=F(x)] 
<- linear_1 successful
 

✓ Mathematica. Time used: 0.02 (sec). Leaf size: 25

ode=D[y[x],{x,2}]+2/x*D[y[x],x]+a^2/x^4*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to c_1\cos \left(\frac{a}{x}\right)-c_2\sin \left(\frac{a}{x}\right)$$

✓ Sympy. Time used: 0.219 (sec). Leaf size: 44

from sympy import * 
x = symbols("x") 
a = symbols("a") 
y = Function("y") 
ode = Eq(a**2*y(x)/x**4 + Derivative(y(x), (x, 2)) + 2*Derivative(y(x), x)/x,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=\frac{\frac{C_1\sqrt{\frac{a}{x}}{J}_{-\frac{1}{2}}\left(\frac{a}{x}\right)}{\sqrt{-\frac{a}{x}}}+C_2{Y}_{-\frac{1}{2}}(-\frac{a}{x})}{\sqrt{x}}$$