2.1.70 Problem 72

Internal problem ID [9240]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 72
Date solved : Friday, April 25, 2025 at 06:03:15 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}(2+x)y^{\prime \prime }+(1+x)y^{\prime }+3y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.463 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (2+x)y^{\prime \prime }+(1+x)y^{\prime }+3y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =2+x\\ \text{(3)}& B& =1+xC& =3\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{x^2-10x-21}{4{(2+x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =x^2-10x-21\\ t& =4{(2+x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{x^2-10x-21}{4{(2+x)}^2}\right)z(x)\end{array}$$

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 2.70: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-2\\ & =0\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=4{(2+x)}^2$. There is a pole at $x=-2$ of order $2$. 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. Since there is a pole of order $2$ then necessary conditions for case two are met. Therefore

$$\begin{array}{rl}L& =[1,2]\end{array}$$

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

Looking at poles of order 2. The partial fractions decomposition of $r$ is

$$r=\frac{1}{4}+\frac{3}{4{(2+x)}^2}-\frac{7}{2(2+x)}$$

For the pole at $x=-2$ let $b$ be the coefficient of $\frac{1}{{(2+x)}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{3}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{3}{2}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{2}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=0$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{0}{2}& & =0\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ is the sum of terms involving $x^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{x}^i\\ \text{(8)}& & =\sum _{i=0}^0{a}_i{x}^i\end{array}$$

Let $a$ be the coefficient of $x^v=x^0$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{1}{2}-\frac{7}{2x}-\frac{9}{2x^2}-\frac{97}{2x^3}-\frac{1291}{4x^4}-\frac{11103}{4x^5}-\frac{98061}{4x^6}-\frac{913053}{4x^7}+\dots \end{array}$$

Comparing Eq. (9) with Eq. (8) shows that

$$a=\frac{1}{2}$$

From Eq. (9) the sum up to $v=0$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^0{a}_i{x}^i\\ \text{(10)}& & =\frac{1}{2}\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^{-1}=\frac{1}{x}$ in $r$ minus the coefficient of same term but in ${([\sqrt{r}{]}_{\mathrm{\infty }})}^2$ where $[\sqrt{r}{]}_{\mathrm{\infty }}$ was found above in Eq (10). Hence

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{1}{4}$$

This shows that the coefficient of $\frac{1}{x}$ in the above is $0$. Now we need to find the coefficient of $\frac{1}{x}$ in $r$. How this is done depends on if $v=0$ or not. Since $v=0$ then starting from $r=\frac{s}{t}$ and doing long division in the form

$$r=Q+\frac{R}{t}$$

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $\frac{1}{x}$ in $r$ will be the coefficient in $R$ of the term in $x$ of degree of $t$ minus one, divided by the leading coefficient in $t$. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2-10x-21}{4x^2+16x+16}\\ & =Q+\frac{R}{4x^2+16x+16}\\ & =\left(\frac{1}{4}\right)+\left(\frac{-14x-25}{4x^2+16x+16}\right)\\ & =\frac{1}{4}+\frac{-14x-25}{4x^2+16x+16}\end{array}$$

Since the degree of $t$ is $2$, then we see that the coefficient of the term $x$ in the remainder $R$ is $-14$. Dividing this by leading coefficient in $t$ which is $4$ gives $-\frac{7}{2}$. Now $b$ can be found.

$$\begin{array}{rl}b& =(-\frac{7}{2})-\left(0\right)\\ & =-\frac{7}{2}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{-\frac{7}{2}}{\frac{1}{2}}-0)& & =-\frac{7}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{-\frac{7}{2}}{\frac{1}{2}}-0)& & =\frac{7}{2}\end{array}$$

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

$$r=\frac{x^2-10x-21}{4{(2+x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$-2$	$2$	$0$	$\frac{3}{2}$	$-\frac{1}{2}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$0$	$\frac{1}{2}$	$-\frac{7}{2}$	$\frac{7}{2}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=\frac{7}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{+}\right)\\ & =\frac{7}{2}-\left(\frac{3}{2}\right)\\ & =2\end{array}$$

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

$$\begin{array}{rl}\omega & =((+)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{+}}{x-c_1})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{3}{2(2+x)}+(-)\left(\frac{1}{2}\right)\\ & =\frac{3}{2(2+x)}-\frac{1}{2}\\ & =-\frac{-1+x}{2(2+x)}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=2$ 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

$$\begin{array}{}\text{(2A)}& p(x)& =x^2+a_{1}x+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(2\right)+2(\frac{3}{2(2+x)}-\frac{1}{2})(2x+a_1)+((-\frac{3}{2{(2+x)}^2})+{(\frac{3}{2(2+x)}-\frac{1}{2})}^2-\left(\frac{x^2-10x-21}{4{(2+x)}^2}\right))& =0\\ \frac{(a_1+4)x+2a_0+a_1+4}{2+x}=0\end{array}$$

Solving for the coefficients $a_i$ in the above using method of undetermined coefficients gives

$$\{a_0=0,a_1=-4\}$$

Substituting these coefficients in $p(x)$ in eq. (2A) results in

$$\begin{array}{rl}p(x)& =x^2-4x\end{array}$$

Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(x^2-4x){\mathrm{e}}^{\int (\frac{3}{2(2+x)}-\frac{1}{2})dx}\\ & =(x^2-4x){\mathrm{e}}^{-\frac{x}{2}+\frac{3\ln (2+x)}{2}}\\ & =x(x-4){(2+x)}^{3/2}{\mathrm{e}}^{-\frac{x}{2}}\end{array}$$

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

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

Which simplifies to

$$y_1={(2+x)}^2{\mathrm{e}}^{-x}x(x-4)$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{1+x}{2+x}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-x+\ln (2+x)}}{{\left(y_1\right)}^2}dx\\ & =y_1(-\frac{{\mathrm{e}}^{-2}{\mathrm{Ei}}_1(-x-2)}{48}-\frac{{\mathrm{e}}^x}{3456(x-4)}-\frac{{\mathrm{e}}^x}{288{(2+x)}^2}-\frac{11{\mathrm{e}}^x}{864(2+x)}-\frac{{\mathrm{e}}^x}{128x})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left({(2+x)}^2{\mathrm{e}}^{-x}x(x-4)\right)+c_2\left({(2+x)}^2{\mathrm{e}}^{-x}x(x-4)(-\frac{{\mathrm{e}}^{-2}{\mathrm{Ei}}_1(-x-2)}{48}-\frac{{\mathrm{e}}^x}{3456(x-4)}-\frac{{\mathrm{e}}^x}{288{(2+x)}^2}-\frac{11{\mathrm{e}}^x}{864(2+x)}-\frac{{\mathrm{e}}^x}{128x})\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.005 (sec). Leaf size: 58

ode:=(x+2)*diff(diff(y(x),x),x)+(1+x)*diff(y(x),x)+3*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1{\mathrm{e}}^{-x}x(x-4){(2+x)}^2+c_2({\mathrm{e}}^{-2-x}x(x-4){(2+x)}^2{\mathrm{Ei}}_1(-2-x)+x^3-x^2-10x-6)$$

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) 
   Group is reducible, not completely reducible 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & (2+x)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(1+x)\left(\frac{d}{dx}y\left(x\right)\right)+3y\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{3y\left(x\right)}{2+x}-\frac{(1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{2+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{(1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{2+x}+\frac{3y\left(x\right)}{2+x}=0\\ ◻& & \text{Check to see if}{x}_0=-2\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{1+x}{2+x},P_3\left(x\right)=\frac{3}{2+x}]\\ & \circ & (2+x)\cdot P_2\left(x\right)\text{is analytic at}x=-2\\ & & ((2+x)\cdot P_2\left(x\right))|{}_{x=-2}=-1\\ & \circ & {(2+x)}^2\cdot P_3\left(x\right)\text{is analytic at}x=-2\\ & & ({(2+x)}^2\cdot P_3\left(x\right))|{}_{x=-2}=0\\ & \circ & x=-2\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=-2\text{is a regular singular point}\\ & & x_0=-2\\ \bullet & & \text{Multiply by denominators}\\ & & (2+x)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(1+x)\left(\frac{d}{dx}y\left(x\right)\right)+3y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-2\text{so that the regular singular point is at}u=0\\ & & u\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(-1+u)\left(\frac{d}{du}y\left(u\right)\right)+3y\left(u\right)=0\\ \bullet & & \text{Assume series solution for}y\left(u\right)\\ & & y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=0..1\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)u^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)u^{k+r}\\ & \circ & \text{Convert}u\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)\text{to series expansion}\\ & & u\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)u^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & u\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)(k+r)u^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(-2+r)u^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(a_{k+1}(k+1+r)(k+r-1)+a_k(k+r+3))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(-2+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,2\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}(k+1+r)(k+r-1)+a_k(k+r+3)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=-\frac{a_k(k+r+3)}{(k+1+r)(k+r-1)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=-\frac{a_k(k+3)}{(k+1)(k-1)}\\ \bullet & & \text{Series not valid for}r=0\text{, division by}0\text{in the recursion relation at}k=1\\ & & a_{k+1}=-\frac{a_k(k+3)}{(k+1)(k-1)}\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+1}=-\frac{a_k(k+5)}{(k+3)(k+1)}\\ \bullet & & \text{Solution for}r=2\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+2},a_{k+1}=-\frac{a_k(k+5)}{(k+3)(k+1)}]\\ \bullet & & \text{Revert the change of variables}u=2+x\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(2+x)}^{k+2},a_{k+1}=-\frac{a_k(k+5)}{(k+3)(k+1)}]\end{array}$$

✓ Mathematica. Time used: 0.542 (sec). Leaf size: 106

ode=(2+x)*D[y[x],{x,2}]+(1+x)*D[y[x],x]+3*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to (x-4)x\exp ({\int }_1^x(\frac{3}{2(K[1]+2)}-\frac{1}{2})dK[1]-\frac{1}{2}{\int }_1^x\frac{K[2]+1}{K[2]+2}dK[2])(c_2{\int }_1^x\frac{\exp (-2{\int }_1^{K[3]}(\frac{3}{2(K[1]+2)}-\frac{1}{2})dK[1])}{(K[3]-4{)}^{2}K[3{]}^2}dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((x + 1)*Derivative(y(x), x) + (x + 2)*Derivative(y(x), (x, 2)) + 3*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False