2.1.72 Problem 74

Internal problem ID [9242]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 74
Date solved : Friday, April 25, 2025 at 06:03:18 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}(2x^2+3x)y^{\prime \prime }+10(1+x)y^{\prime }+8y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.385 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (2x^2+3x)y^{\prime \prime }+(10x+10)y^{\prime }+8y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =2x^2+3x\\ \text{(3)}& B& =10x+10C& =8\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-x^2+6x+10}{{(2x^2+3x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-x^2+6x+10\\ t& ={(2x^2+3x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-x^2+6x+10}{{(2x^2+3x)}^2}\right)z(x)\end{array}$$

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 2.72: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-2\\ & =2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t={(2x^2+3x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=-\frac{3}{2}$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $2$ 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 $2$ 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

$$r=-\frac{5}{36{(x+\frac{3}{2})}^2}+\frac{22}{27(x+\frac{3}{2})}-\frac{22}{27x}+\frac{10}{9x^2}$$

For the pole at $x=0$ let $b$ be the coefficient of $\frac{1}{x^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{10}{9}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{5}{3}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{2}{3}\end{array}$$

For the pole at $x=-\frac{3}{2}$ let $b$ be the coefficient of $\frac{1}{{(x+\frac{3}{2})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{5}{36}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{5}{6}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =\frac{1}{6}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is 2 then $[\sqrt{r}{]}_{\mathrm{\infty }}=0$. Let $b$ be the coefficient of $\frac{1}{x^2}$ in the Laurent series expansion of $r$ at $\mathrm{\infty }$. which can be found by dividing the leading coefficient of $s$ by the leading coefficient of $t$ from

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{-x^2+6x+10}{{(2x^2+3x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=-\frac{1}{4}$. Hence

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

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

$$r=\frac{-x^2+6x+10}{{(2x^2+3x)}^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{5}{3}$	$-\frac{2}{3}$
$-\frac{3}{2}$	$2$	$0$	$\frac{5}{6}$	$\frac{1}{6}$

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

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 ${\alpha }_{\mathrm{\infty }}^{-}=\frac{1}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})\\ & =\frac{1}{2}-(-\frac{1}{2})\\ & =1\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+((-)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{-}}{x-c_2})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{2}{3x}+\frac{1}{6x+9}+(-)\left(0\right)\\ & =-\frac{2}{3x}+\frac{1}{6x+9}\\ & =-\frac{x+2}{x(2x+3)}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=1$ 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

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

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{2}{3x}+\frac{1}{6x+9})\left(1\right)+((\frac{2}{3x^2}-\frac{1}{6{(x+\frac{3}{2})}^2})+{(-\frac{2}{3x}+\frac{1}{6x+9})}^2-\left(\frac{-x^2+6x+10}{{(2x^2+3x)}^2}\right))& =0\\ \frac{-4+2a_0}{x(2x+3)}=0\end{array}$$

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

$$\{a_0=2\}$$

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

$$\begin{array}{rl}p(x)& =x+2\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(x+2){\mathrm{e}}^{\int (-\frac{2}{3x}+\frac{1}{6x+9})dx}\\ & =(x+2){\mathrm{e}}^{-\frac{2\ln \left(x\right)}{3}+\frac{\ln (2x+3)}{6}}\\ & =\frac{(x+2){(2x+3)}^{1/6}}{x^{2/3}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{10x+10}{2x^2+3x}dx}\\ & =z_1{e}^{-\frac{5\ln \left(x\right)}{3}-\frac{5\ln (2x+3)}{6}}\\ & =z_1\left(\frac{1}{x^{5/3}{(2x+3)}^{5/6}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{x+2}{x^{7/3}{(2x+3)}^{2/3}}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{10x+10}{2x^2+3x}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{10\ln \left(x\right)}{3}-\frac{5\ln (2x+3)}{3}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{{\mathrm{e}}^{-\frac{10\ln \left(x\right)}{3}-\frac{5\ln (2x+3)}{3}}{x}^{14/3}{(2x+3)}^{4/3}}{{(x+2)}^2}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{x+2}{x^{7/3}{(2x+3)}^{2/3}}\right)+c_2\left(\frac{x+2}{x^{7/3}{(2x+3)}^{2/3}}(\int \frac{{\mathrm{e}}^{-\frac{10\ln \left(x\right)}{3}-\frac{5\ln (2x+3)}{3}}{x}^{14/3}{(2x+3)}^{4/3}}{{(x+2)}^2}dx)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.135 (sec). Leaf size: 31

ode:=(2*x^2+3*x)*diff(diff(y(x),x),x)+10*(1+x)*diff(y(x),x)+8*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{c_1(2+x)}{{(1+\frac{2x}{3})}^{2/3}{x}^{7/3}}+c_2\mathrm{hypergeom}([2,2],\left[\frac{10}{3}\right],-\frac{2x}{3})$$

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 
   Solution has integrals. Trying a special function solution free of integrals\ 
... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Kummer 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
         <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved -\ 
 returning with no uncomputed integrals 
   <- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & (2x^2+3x)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+10(1+x)\left(\frac{d}{dx}y\left(x\right)\right)+8y\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{8y\left(x\right)}{x(2x+3)}-\frac{10(1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x(2x+3)}\\ \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{10(1+x)\left(\frac{d}{dx}y\left(x\right)\right)}{x(2x+3)}+\frac{8y\left(x\right)}{x(2x+3)}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{10(1+x)}{x(2x+3)},P_3\left(x\right)=\frac{8}{x(2x+3)}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=\frac{10}{3}\\ & \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\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & x(2x+3)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+(10x+10)\left(\frac{d}{dx}y\left(x\right)\right)+8y\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}^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=0..1\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)x^{k+r}\\ & \circ & \text{Convert}{x}^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=1..2\\ & & x^m\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-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & x^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(7+3r)x^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(a_{k+1}(k+1+r)(3k+10+3r)+2a_k{(k+r+2)}^2)x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(7+3r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,-\frac{7}{3}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}(k+1+r)(3k+10+3r)+2a_k{(k+r+2)}^2=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=-\frac{2a_k{(k+r+2)}^2}{(k+1+r)(3k+10+3r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=-\frac{2a_k{(k+2)}^2}{(k+1)(3k+10)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+1}=-\frac{2a_k{(k+2)}^2}{(k+1)(3k+10)}]\\ \bullet & & \text{Recursion relation for}r=-\frac{7}{3}\\ & & a_{k+1}=-\frac{2a_k{(k-\frac{1}{3})}^2}{(k-\frac{4}{3})(3k+3)}\\ \bullet & & \text{Solution for}r=-\frac{7}{3}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k-\frac{7}{3}},a_{k+1}=-\frac{2a_k{(k-\frac{1}{3})}^2}{(k-\frac{4}{3})(3k+3)}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k-\frac{7}{3}}),a_{k+1}=-\frac{2a_k{(k+2)}^2}{(k+1)(3k+10)},b_{k+1}=-\frac{2b_k{(k-\frac{1}{3})}^2}{(k-\frac{4}{3})(3k+3)}]\end{array}$$

✓ Mathematica. Time used: 0.439 (sec). Leaf size: 118

ode=(3*x+2*x^2)*D[y[x],{x,2}]+10*(1+x)*D[y[x],x]+8*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to (x+2)\exp ({\int }_1^x-\frac{K[1]+2}{2K[1{]}^2+3K[1]}dK[1]-\frac{1}{2}{\int }_1^x\frac{10(K[2]+1)}{K[2](2K[2]+3)}dK[2])(c_2{\int }_1^x\frac{\exp (-2{\int }_1^{K[3]}-\frac{K[1]+2}{2K[1{]}^2+3K[1]}dK[1])}{(K[3]+2{)}^2}dK[3]+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((10*x + 10)*Derivative(y(x), x) + (2*x**2 + 3*x)*Derivative(y(x), (x, 2)) + 8*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False