2.1.62 Problem 64

Internal problem ID [9232]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 64
Date solved : Friday, April 25, 2025 at 06:00:26 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}(2x^2+4x+5)y^{\prime \prime }-20(x+1)y^{\prime }+60y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.771 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (2x^2+4x+5)y^{\prime \prime }+(-20x-20)y^{\prime }+60y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =2x^2+4x+5\\ \text{(3)}& B& =-20x-20C& =60\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-210}{{(2x^2+4x+5)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-210\\ t& ={(2x^2+4x+5)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(-\frac{210}{{(2x^2+4x+5)}^2})z(x)\end{array}$$

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 2.62: 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 $t={(2x^2+4x+5)}^2$. There is a pole at $x=-1+\frac{i\sqrt{6}}{2}$ of order $2$. There is a pole at $x=-1-\frac{i\sqrt{6}}{2}$ of order $2$. 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. Since there is a pole of order $2$ then necessary conditions for case two are met. Since pole order is not larger than $2$ and the order at $\mathrm{\infty }$ is $4$ then the necessary conditions for case three are met. Therefore

$$\begin{array}{rl}L& =[1,2,4,6,12]\end{array}$$

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

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

$$r=\frac{35}{4{(x+1-\frac{i\sqrt{6}}{2})}^2}+\frac{35}{4{(x+1+\frac{i\sqrt{6}}{2})}^2}+\frac{35i\sqrt{6}}{12(x+1-\frac{i\sqrt{6}}{2})}-\frac{35i\sqrt{6}}{12(x+1+\frac{i\sqrt{6}}{2})}$$

For the pole at $x=-1+\frac{i\sqrt{6}}{2}$ let $b$ be the coefficient of $\frac{1}{{(x+1-\frac{i\sqrt{6}}{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{35}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{7}{2}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{5}{2}\end{array}$$

For the pole at $x=-1-\frac{i\sqrt{6}}{2}$ let $b$ be the coefficient of $\frac{1}{{(x+1+\frac{i\sqrt{6}}{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{35}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{7}{2}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{5}{2}\end{array}$$

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

$$r=-\frac{210}{{(2x^2+4x+5)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$-1+\frac{i\sqrt{6}}{2}$	$2$	$0$	$\frac{7}{2}$	$-\frac{5}{2}$
$-1-\frac{i\sqrt{6}}{2}$	$2$	$0$	$\frac{7}{2}$	$-\frac{5}{2}$

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

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})\\ & =1-\left(1\right)\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+((+)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{+}}{x-c_2})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{5}{2(x+1-\frac{i\sqrt{6}}{2})}+\frac{7}{2(x+1+\frac{i\sqrt{6}}{2})}+(-)\left(0\right)\\ & =-\frac{5}{2(x+1-\frac{i\sqrt{6}}{2})}+\frac{7}{2(x+1+\frac{i\sqrt{6}}{2})}\\ & =\frac{-6i\sqrt{6}+2x+2}{2x^2+4x+5}\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(-\frac{5}{2(x+1-\frac{i\sqrt{6}}{2})}+\frac{7}{2(x+1+\frac{i\sqrt{6}}{2})})\left(0\right)+((\frac{5}{2{(x+1-\frac{i\sqrt{6}}{2})}^2}-\frac{7}{2{(x+1+\frac{i\sqrt{6}}{2})}^2})+{(-\frac{5}{2(x+1-\frac{i\sqrt{6}}{2})}+\frac{7}{2(x+1+\frac{i\sqrt{6}}{2})})}^2-(-\frac{210}{{(2x^2+4x+5)}^2}))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & ={\mathrm{e}}^{\int (-\frac{5}{2(x+1-\frac{i\sqrt{6}}{2})}+\frac{7}{2(x+1+\frac{i\sqrt{6}}{2})})dx}\\ & =\frac{27\sqrt{2}{(2x^2+4x+5)}^{7/2}}{{(3+i(x+1)\sqrt{6})}^6}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-20x-20}{2x^2+4x+5}dx}\\ & =z_1{e}^{\frac{5\ln (2x^2+4x+5)}{2}}\\ & =z_1\left({(2x^2+4x+5)}^{5/2}\right)\end{array}$$

Which simplifies to

$$y_1=-\frac{{(2x^2+4x+5)}^6\sqrt{2}}{27{(i-\frac{(x+1)\sqrt{2}\sqrt{3}}{3})}^6}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-20x-20}{2x^2+4x+5}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{5\ln (2x^2+4x+5)}}{{\left(y_1\right)}^2}dx\\ & =y_1(-\frac{\frac{1}{2}{x}^5+\frac{5}{2}{x}^4+\frac{5}{2}{x}^3-\frac{5}{2}{x}^2-\frac{31}{8}x-\frac{7}{8}}{2{(x+1+\frac{i\sqrt{6}}{2})}^6})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1(-\frac{{(2x^2+4x+5)}^6\sqrt{2}}{27{(i-\frac{(x+1)\sqrt{2}\sqrt{3}}{3})}^6})+c_2(-\frac{{(2x^2+4x+5)}^6\sqrt{2}}{27{(i-\frac{(x+1)\sqrt{2}\sqrt{3}}{3})}^6}(-\frac{\frac{1}{2}{x}^5+\frac{5}{2}{x}^4+\frac{5}{2}{x}^3-\frac{5}{2}{x}^2-\frac{31}{8}x-\frac{7}{8}}{2{(x+1+\frac{i\sqrt{6}}{2})}^6}))\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.003 (sec). Leaf size: 65

ode:=(2*x^2+4*x+5)*diff(diff(y(x),x),x)-20*(1+x)*diff(y(x),x)+60*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_2{x}^6+c_1{x}^5+\frac{5(2c_1-15c_2)x^4}{2}+5(c_1-20c_2)x^3+\frac{5(-4c_1-45c_2)x^2}{4}+\frac{(-31c_1+120c_2)x}{4}-\frac{7c_1}{4}+\frac{155c_2}{8}$$

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) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & (2x^2+4x+5)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-20(x+1)\left(\frac{d}{dx}y\left(x\right)\right)+60y\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{60y\left(x\right)}{2x^2+4x+5}+\frac{20(x+1)\left(\frac{d}{dx}y\left(x\right)\right)}{2x^2+4x+5}\\ \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{20(x+1)\left(\frac{d}{dx}y\left(x\right)\right)}{2x^2+4x+5}+\frac{60y\left(x\right)}{2x^2+4x+5}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=-\frac{20(x+1)}{2x^2+4x+5},P_3\left(x\right)=\frac{60}{2x^2+4x+5}]\\ & \circ & (x+1+\frac{\mathrm{I}\sqrt{6}}{2})\cdot P_2\left(x\right)\text{is analytic at}x=-1-\frac{\mathrm{I}\sqrt{6}}{2}\\ & & ((x+1+\frac{\mathrm{I}\sqrt{6}}{2})\cdot P_2\left(x\right))|{}_{x=-1-\frac{\mathrm{I}\sqrt{6}}{2}}=0\\ & \circ & {(x+1+\frac{\mathrm{I}\sqrt{6}}{2})}^2\cdot P_3\left(x\right)\text{is analytic at}x=-1-\frac{\mathrm{I}\sqrt{6}}{2}\\ & & ({(x+1+\frac{\mathrm{I}\sqrt{6}}{2})}^2\cdot P_3\left(x\right))|{}_{x=-1-\frac{\mathrm{I}\sqrt{6}}{2}}=0\\ & \circ & x=-1-\frac{\mathrm{I}\sqrt{6}}{2}\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=-1-\frac{\mathrm{I}\sqrt{6}}{2}\\ \bullet & & \text{Multiply by denominators}\\ & & (2x^2+4x+5)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(-20x-20)\left(\frac{d}{dx}y\left(x\right)\right)+60y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-1-\frac{\mathrm{I}\sqrt{6}}{2}\text{so that the regular singular point is at}u=0\\ & & (2u^2-2\mathrm{I}u\sqrt{6})\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(-20u+10\mathrm{I}\sqrt{6})\left(\frac{d}{du}y\left(u\right)\right)+60y\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}^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=1..2\\ & & u^m\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-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & u^m\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)u^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & -2\mathrm{I}\sqrt{6}(r-6)r{a}_0{u}^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(-2\mathrm{I}\sqrt{6}(k+r-5)(k+1+r)a_{k+1}+2a_k(k+r-5)(k+r-6))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -2\mathrm{I}\sqrt{6}(r-6)r=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,6\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & -2(\mathrm{I}(k+1+r)a_{k+1}\sqrt{6}-a_k(k+r-6))(k+r-5)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{-\frac{\mathrm{I}}{6}{a}_k(k+r-6)\sqrt{6}}{k+1+r}\\ \bullet & & \text{Recursion relation for}r=0\text{; series terminates at}k=6\\ & & a_{k+1}=\frac{-\frac{\mathrm{I}}{6}{a}_k(k-6)\sqrt{6}}{k+1}\\ \bullet & & \text{Recursion relation that defines the terminating series solution of the ODE for}r=0\\ & & [y\left(u\right)=\stackrel{5}{\sum _{k=0}}{a}_k{u}^k,a_{k+1}=\frac{-\frac{\mathrm{I}}{6}{a}_k(k-6)\sqrt{6}}{k+1}]\\ \bullet & & \text{Revert the change of variables}u=x+1+\frac{\mathrm{I}\sqrt{6}}{2}\\ & & [y\left(x\right)=\stackrel{5}{\sum _{k=0}}{a}_k{(x+1+\frac{\mathrm{I}\sqrt{6}}{2})}^k,a_{k+1}=\frac{-\frac{\mathrm{I}}{6}{a}_k(k-6)\sqrt{6}}{k+1}]\\ \bullet & & \text{Recursion relation for}r=6\\ & & a_{k+1}=\frac{-\frac{\mathrm{I}}{6}{a}_{k}k\sqrt{6}}{k+7}\\ \bullet & & \text{Solution for}r=6\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+6},a_{k+1}=\frac{-\frac{\mathrm{I}}{6}{a}_{k}k\sqrt{6}}{k+7}]\\ \bullet & & \text{Revert the change of variables}u=x+1+\frac{\mathrm{I}\sqrt{6}}{2}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+1+\frac{\mathrm{I}\sqrt{6}}{2})}^{k+6},a_{k+1}=\frac{-\frac{\mathrm{I}}{6}{a}_{k}k\sqrt{6}}{k+7}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{5}{\sum _{k=0}}{a}_k{(x+1+\frac{\mathrm{I}\sqrt{6}}{2})}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(x+1+\frac{\mathrm{I}\sqrt{6}}{2})}^{k+6}),a_{k+1}=\frac{-\frac{\mathrm{I}}{6}{a}_k(k-6)\sqrt{6}}{k+1},b_{k+1}=\frac{-\frac{\mathrm{I}}{6}{b}_{k}k\sqrt{6}}{k+7}]\end{array}$$

✓ Mathematica. Time used: 0.582 (sec). Leaf size: 108

ode=(2*x^2+4*x+5)*D[y[x],{x,2}]-20*(x+1)*D[y[x],x]+60*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to {(2x^2+4x+5)}^{5/2}\exp ({\int }_1^x\frac{2(K[1]+3i\sqrt{6}+1)}{2K[1](K[1]+2)+5}dK[1])(c_2{\int }_1^x\exp (-2{\int }_1^{K[2]}\frac{2(K[1]+3i\sqrt{6}+1)}{2K[1](K[1]+2)+5}dK[1])dK[2]+c_1)$$

✓ Sympy. Time used: 1.102 (sec). Leaf size: 3

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((-20*x - 20)*Derivative(y(x), x) + (2*x**2 + 4*x + 5)*Derivative(y(x), (x, 2)) + 60*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=O\left(1\right)$$