2.2.37 Problem 37

Internal problem ID [9159]
Book : Second order enumerated odes
Section : section 2
Problem number : 37
Date solved : Friday, April 25, 2025 at 05:58:31 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

Time used: 0.147 (sec)

Solve

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

In normal form the given ode is written as

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

Where

$$\begin{array}{rl}p\left(x\right)& =\frac{2}{x}\\ q\left(x\right)& =-1\end{array}$$

Calculating the Liouville ode invariant $Q$ given by

$$\begin{array}{rl}Q& =q-\frac{p^{\prime }}{2}-\frac{p^2}{4}\\ & =-1-\frac{{\left(\frac{2}{x}\right)}^{\prime }}{2}-\frac{{\left(\frac{2}{x}\right)}^2}{4}\\ & =-1-\frac{(-\frac{2}{x^2})}{2}-\frac{\left(\frac{4}{x^2}\right)}{4}\\ & =-1-(-\frac{1}{x^2})-\frac{1}{x^2}\\ & =-1\end{array}$$

Since the Liouville ode invariant does not depend on the independent variable $x$ then the transformation

$$\begin{array}{}\text{(3)}& y=v\left(x\right)z\left(x\right)\end{array}$$

is used to change the original ode to a constant coefficients ode in $v$. In (3) the term $z\left(x\right)$ is given by

$$\begin{array}{rl}z\left(x\right)& ={\mathrm{e}}^{-\int \frac{p\left(x\right)}{2}dx}\\ & =e^{-\int \frac{\frac{2}{x}}{2}}\\ \text{(5)}& & =\frac{1}{x}\end{array}$$

Hence (3) becomes

$$\begin{array}{}\text{(4)}& y=\frac{v\left(x\right)}{x}\end{array}$$

Applying this change of variable to the original ode results in

$$\begin{array}{r}{v}^{\prime \prime }\left(x\right)-v\left(x\right)=0\end{array}$$

Which is now solved for $v\left(x\right)$.

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{v}^{″}(x)+B{v}^{\prime }(x)+Cv(x)=0$$

Where in the above $A=1,B=0,C=-1$. Let the solution be $v=e^{\lambda x}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{x\lambda }-{\mathrm{e}}^{x\lambda }=0\end{array}$$

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

$$\begin{array}{}\text{(2)}& {\lambda }^2-1=0\end{array}$$

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 $A=1,B=0,C=-1$ into the above gives

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

Hence

$$\begin{array}{rl}{\lambda }_1& =+1\\ {\lambda }_2& =-1\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =1\\ {\lambda }_2& =-1\end{array}$$

Since roots are real and distinct, then the solution is

$$\begin{array}{rl}v& =c_1{e}^{{\lambda }_{1}x}+c_2{e}^{{\lambda }_{2}x}\\ v& =c_1{e}^{\left(1\right)x}+c_2{e}^{(-1)x}\end{array}$$

Or

$$v=c_1{\mathrm{e}}^x+c_2{\mathrm{e}}^{-x}$$

Will add steps showing solving for IC soon.

Now that $v$ is known, then

$$\begin{array}{rl}y\left(x\right)& =vz\left(x\right)\\ \text{(7)}& & =(c_1{\mathrm{e}}^x+c_2{\mathrm{e}}^{-x})\left(z\left(x\right)\right)\end{array}$$

But from (5)

$$\begin{array}{rl}z\left(x\right)& =\frac{1}{x}\end{array}$$

Hence (7) becomes

$$\begin{array}{r}y\left(x\right)=\frac{c_1{\mathrm{e}}^x+c_2{\mathrm{e}}^{-x}}{x}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

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

Solved as second order Bessel ode

Time used: 0.106 (sec)

Solve

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

Writing the ode as

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

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

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

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

$$\begin{array}{r}y=\frac{i{c}_1\sqrt{2}\sinh \left(x\right)}{\sqrt{x}\sqrt{\pi }\sqrt{ix}}-\frac{c_2\sqrt{2}\cosh \left(x\right)}{\sqrt{x}\sqrt{\pi }\sqrt{ix}}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{i{c}_1\sqrt{2}\sinh \left(x\right)}{\sqrt{x}\sqrt{\pi }\sqrt{ix}}-\frac{c_2\sqrt{2}\cosh \left(x\right)}{\sqrt{x}\sqrt{\pi }\sqrt{ix}}\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.076 (sec)

Solve

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

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =x\\ \text{(3)}& B& =2C& =-x\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{1}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =1\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =z\left(x\right)\end{array}$$

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 2.46: Necessary conditions for each Kovacic case

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

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

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $0$ then the necessary conditions for case one are met. Therefore

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

Since $r=1$ is not a function of $x$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(x)={\mathrm{e}}^{-x}$$

Using the above, the solution for the original ode can now be found. The first solution to the original ode in $y$ is found from

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

Which simplifies to

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

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{c_1{\mathrm{e}}^{-x}}{x}+\frac{c_2{\mathrm{e}}^x}{2x}\end{array}$$

✓ Maple. Time used: 0.005 (sec). Leaf size: 17

ode:=x*diff(diff(y(x),x),x)+2*diff(y(x),x)-x*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{c_1\sinh \left(x\right)+c_2\cosh \left(x\right)}{x}$$

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)] 
checking if the LODE is missing y 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+2\frac{d}{dx}y\left(x\right)-xy\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)=y\left(x\right)-\frac{2\left(\frac{d}{dx}y\left(x\right)\right)}{x}\\ \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{2\left(\frac{d}{dx}y\left(x\right)\right)}{x}-y\left(x\right)=0\\ ◻& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{2}{x},P_3\left(x\right)=-1]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=2\\ & \circ & x^2\cdot P_3\left(x\right)\text{is analytic at}x=0\\ & & (x^2\cdot P_3\left(x\right))|{}_{x=0}=0\\ & \circ & x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & x\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+2\frac{d}{dx}y\left(x\right)-xy\left(x\right)=0\\ \bullet & & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}x\cdot y\left(x\right)\text{to series expansion}\\ & & x\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r+1}\\ & \circ & \text{Shift index using}k->k-1\\ & & x\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=1}}{a}_{k-1}{x}^{k+r}\\ & \circ & \text{Convert}\frac{d}{dx}y\left(x\right)\text{to series expansion}\\ & & \frac{d}{dx}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & \frac{d}{dx}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+r+1)x^{k+r}\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)x^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & x\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+r+1)(k+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(1+r)x^{-1+r}+a_1(1+r)(2+r)x^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_{k+1}(k+r+1)(k+2+r)-a_{k-1})x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(1+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{-1,0\}\\ \bullet & & \text{Each term must be 0}\\ & & a_1(1+r)(2+r)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}(k+r+1)(k+2+r)-a_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & a_{k+2}(k+2+r)(k+3+r)-a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=\frac{a_k}{(k+2+r)(k+3+r)}\\ \bullet & & \text{Recursion relation for}r=-1\\ & & a_{k+2}=\frac{a_k}{(k+1)(k+2)}\\ \bullet & & \text{Solution for}r=-1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-1},a_{k+2}=\frac{a_k}{(k+1)(k+2)},0=0]\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=\frac{a_k}{(k+2)(k+3)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+2}=\frac{a_k}{(k+2)(k+3)},2a_1=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-1})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^k),a_{k+2}=\frac{a_k}{(k+1)(k+2)},0=0,b_{k+2}=\frac{b_k}{(k+2)(k+3)},2b_1=0]\end{array}$$

✓ Mathematica. Time used: 0.03 (sec). Leaf size: 28

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

$$y(x)\to \frac{2c_1{e}^{-x}+c_2{e}^x}{2x}$$

✓ Sympy. Time used: 0.210 (sec). Leaf size: 24

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

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