2.1.200 Problem 202

Internal problem ID [9372]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 202
Date solved : Wednesday, March 05, 2025 at 07:48:14 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 0.322 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& ={(1+t)}^{2}C& =-4-4t\end{array}$$

Applying the Liouville transformation on the dependent variable gives

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

Then (2) becomes

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{t^4+4t^3+6t^2+24t+21}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =t^4+4t^3+6t^2+24t+21\\ t& =4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =(\frac{21}{4}+6t+\frac{1}{4}{t}^4+t^3+\frac{3}{2}{t}^2)z(t)\end{array}$$

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

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

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 2.200: Necessary conditions for each Kovacic case

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

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

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 $-4$ then the necessary conditions for case one are met. Therefore

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

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

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=-4$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{4}{2}& & =2\end{array}$$

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

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{t}^i\\ \text{(8)}& & =\sum _{i=0}^2{a}_i{t}^i\end{array}$$

Let $a$ be the coefficient of $t^v=t^2$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{t^2}{2}+t+\frac{1}{2}+\frac{5}{t}-\frac{5}{t^2}+\frac{5}{t^3}-\frac{30}{t^4}+\frac{105}{t^5}+\dots \end{array}$$

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

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

From Eq. (9) the sum up to $v=2$ gives

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

Now we need to find $b$, where $b$ be the coefficient of $t^{v-1}=t^1=t$ 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

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{1}{4}{t}^4+t^3+\frac{3}{2}{t}^2+t+\frac{1}{4}$$

This shows that the coefficient of $t$ in the above is $1$. Now we need to find the coefficient of $t$ in $r$. How this is done depends on if $v=0$ or not. Since $v=2$ which is not zero, then starting $r=\frac{s}{t}$, we do long division and write this in the form

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

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $t$ in $r$ will be the coefficient this term in the quotient. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{t^4+4t^3+6t^2+24t+21}{4}\\ & =Q+\frac{R}{4}\\ & =(\frac{21}{4}+6t+\frac{1}{4}{t}^4+t^3+\frac{3}{2}{t}^2)+\left(0\right)\\ & =\frac{21}{4}+6t+\frac{1}{4}{t}^4+t^3+\frac{3}{2}{t}^2\end{array}$$

We see that the coefficient of the term $\frac{1}{t}$ in the quotient is $6$. Now $b$ can be found.

$$\begin{array}{rl}b& =\left(6\right)-\left(1\right)\\ & =5\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{1}{2}{t}^2+t+\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{5}{\frac{1}{2}}-2)& & =4\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{5}{\frac{1}{2}}-2)& & =-6\end{array}$$

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

$$r=\frac{21}{4}+6t+\frac{1}{4}{t}^4+t^3+\frac{3}{2}{t}^2$$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$-4$	$\frac{1}{2}{t}^2+t+\frac{1}{2}$	$4$	$-6$

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 ${\alpha }_{\mathrm{\infty }}^{+}=4$, and since there are no poles, then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}\\ & =4\end{array}$$

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

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

Substituting the above values in the above results in

$$\begin{array}{rl}\omega & =(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =0+(\frac{1}{2}{t}^2+t+\frac{1}{2})\\ & =\frac{1}{2}{t}^2+t+\frac{1}{2}\\ & =\frac{{(1+t)}^2}{2}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(t)$ of degree $d=4$ to solve the ode. The polynomial $p(t)$ needs to satisfy the equation

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

Let

$$\begin{array}{}\text{(2A)}& p(t)& =t^4+a_3{t}^3+a_2{t}^2+a_{1}t+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}(12t^2+6t{a}_3+2a_2)+2(\frac{1}{2}{t}^2+t+\frac{1}{2})(4t^3+3t^2{a}_3+2t{a}_2+a_1)+((1+t)+{(\frac{1}{2}{t}^2+t+\frac{1}{2})}^2-(\frac{21}{4}+6t+\frac{1}{4}{t}^4+t^3+\frac{3}{2}{t}^2))& =0\\ (-a_3+4)t^4+2(2-a_2+a_3)t^3+3(4-a_1+a_3)t^2+2(-2a_0-a_1+a_2+3a_3)t-4a_0+a_1+2a_2=0\end{array}$$

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

$$\{a_0=5,a_1=8,a_2=6,a_3=4\}$$

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

$$\begin{array}{rl}p(t)& =t^4+4t^3+6t^2+8t+5\end{array}$$

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

$$\begin{array}{rl}{z}_1(t)& =p{e}^{\int \omega dt}\\ & =(t^4+4t^3+6t^2+8t+5){\mathrm{e}}^{\int (\frac{1}{2}{t}^2+t+\frac{1}{2})dt}\\ & =(t^4+4t^3+6t^2+8t+5){\mathrm{e}}^{\frac{{(1+t)}^3}{6}}\\ & =(1+t)(t^3+3t^2+3t+5){\mathrm{e}}^{\frac{{(1+t)}^3}{6}}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{{(1+t)}^2}{1}dt}\\ & =z_1{e}^{-\frac{{(1+t)}^3}{6}}\\ & =z_1\left({\mathrm{e}}^{-\frac{{(1+t)}^3}{6}}\right)\end{array}$$

Which simplifies to

$$y_1=(1+t)(t^3+3t^2+3t+5)$$

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

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

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{{(1+t)}^2}{1}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{-\frac{{(1+t)}^3}{3}}}{{\left(y_1\right)}^2}dt\\ & =y_1(\int \frac{{\mathrm{e}}^{-\frac{{(1+t)}^3}{3}}}{{(1+t)}^2{(t^3+3t^2+3t+5)}^2}dt)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left((1+t)(t^3+3t^2+3t+5)\right)+c_2\left((1+t)(t^3+3t^2+3t+5)(\int \frac{{\mathrm{e}}^{-\frac{{(1+t)}^3}{3}}}{{(1+t)}^2{(t^3+3t^2+3t+5)}^2}dt)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 1.993 (sec). Leaf size: 60

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

$$y=(t+1)(t^3+3t^2+3t+5)(c_2(\int \frac{{\mathrm{e}}^{-\frac{t(t^2+3t+3)}{3}}}{{(t+1)}^2{(t^3+3t^2+3t+5)}^2}dt)+c_1)$$

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 
      -> Mathieu 
         -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
      <- Heun successful: received ODE is equivalent to the  HeunT  ODE, case  c = 0 
   Special function solution also has integrals. Returning default Liouvillian solution. 
<- Kovacics algorithm successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d^2}{d{t}^2}y\left(t\right)+(t^2+2t+1)\left(\frac{d}{dt}y\left(t\right)\right)-(4t+4)y\left(t\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{t}^2}y\left(t\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d^2}{d{t}^2}y\left(t\right)=-(t^2+2t+1)\left(\frac{d}{dt}y\left(t\right)\right)+(4t+4)y\left(t\right)\\ \bullet & & \text{Group terms with}y\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d^2}{d{t}^2}y\left(t\right)+(t^2+2t+1)\left(\frac{d}{dt}y\left(t\right)\right)+(-4t-4)y\left(t\right)=0\\ \bullet & & \text{Assume series solution for}y\left(t\right)\\ & & y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^k\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{t}^m\cdot y\left(t\right)\text{to series expansion for}m=0..1\\ & & t^m\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,-m)}}{a}_k{t}^{k+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & t^m\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,-m)+m}}{a}_{k-m}{t}^k\\ & \circ & \text{Convert}{t}^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)\text{to series expansion for}m=0..2\\ & & t^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,1-m)}}{a}_{k}k{t}^{k-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & t^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,1-m)+m-1}}{a}_{k+1-m}(k+1-m)t^k\\ & \circ & \text{Convert}\frac{d^2}{d{t}^2}y\left(t\right)\text{to series expansion}\\ & & \frac{d^2}{d{t}^2}y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=2}}{a}_{k}k(k-1)t^{k-2}\\ & \circ & \text{Shift index using}k->k+2\\ & & \frac{d^2}{d{t}^2}y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_{k+2}(k+2)(k+1)t^k\\ & & \text{Rewrite ODE with series expansions}\\ & & 2a_2+a_1-4a_0+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_{k+2}(k+2)(k+1)+a_{k+1}(k+1)+2a_k(k-2)+a_{k-1}(k-5))t^k)=0\\ \bullet & & \text{Each term must be 0}\\ & & 2a_2+a_1-4a_0=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & k^2{a}_{k+2}+(2a_k+a_{k-1}+a_{k+1}+3a_{k+2})k-4a_k-5a_{k-1}+a_{k+1}+2a_{k+2}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & {(k+1)}^2{a}_{k+3}+(2a_{k+1}+a_k+a_{k+2}+3a_{k+3})(k+1)-4a_{k+1}-5a_k+a_{k+2}+2a_{k+3}=0\\ \bullet & & \text{Recursion relation that defines the series solution to the ODE}\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^k,a_{k+3}=-\frac{a_{k}k+2a_{k+1}k+k{a}_{k+2}-4a_k-2a_{k+1}+2a_{k+2}}{k^2+5k+6},2a_2+a_1-4a_0=0]\end{array}$$

✓ Mathematica. Time used: 4.118 (sec). Leaf size: 78

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

$$y(t)\to (t+1)(t^3+3t^2+3t+5)(c_2{\int }_1^t\frac{e^{-\frac{1}{3}K[1](K[1{]}^2+3K[1]+3)}}{(K[1]+1{)}^2{(K[1{]}^3+3K[1{]}^2+3K[1]+5)}^2}dK[1]+c_1)$$

✓ Sympy. Time used: 0.890 (sec). Leaf size: 32

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq((-4*t - 4)*y(t) + (t**2 + 2*t + 1)*Derivative(y(t), t) + Derivative(y(t), (t, 2)),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=C_2(2t^2+1)+C_{1}t(\frac{t^3}{8}+\frac{t^2}{2}-\frac{t}{2}+1)+O\left(t^6\right)$$