2.1.51 Problem 51

Internal problem ID [8763]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 51
Date solved : Sunday, March 30, 2025 at 01:30:41 PM
CAS classification : [[_2nd_order, _missing_y]]

Solved as second order missing y ode

Time used: 0.425 (sec)

Solve

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

This is second order ode with missing dependent variable $y$. Let

$$\begin{array}{rl}u(t)& =y^{\prime }\end{array}$$

Then

$$\begin{array}{rl}{u}^{\prime }(t)& =y^{\prime \prime }\end{array}$$

Hence the ode becomes

$$\begin{array}{r}{t}^2{u}^{\prime }\left(t\right)-2u\left(t\right)=0\end{array}$$

Which is now solved for $u(t)$ as first order ode.

The ode

$$\begin{array}{}\text{(1)}& u^{\prime }=\frac{2u}{t^2}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{u}^{\prime }& =\frac{2u}{t^2}\\ & =f(t)g(u)\end{array}$$

Where

$$\begin{array}{rl}f(t)& =\frac{2}{t^2}\\ g(u)& =u\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(u)}du& =\int f(t)dt\\ \int \frac{1}{u}du& =\int \frac{2}{t^2}dt\end{array}$$

$$\ln \left(u\right)=-\frac{2}{t}+c_1$$

We now need to find the singular solutions, these are found by finding for what values $g(u)$ is zero, since we had to divide by this above. Solving $g(u)=0$ or

$$u=0$$

for $u$ gives

$$\begin{array}{rl}u& =0\end{array}$$

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

$$\begin{array}{rl}\ln \left(u\right)& =-\frac{2}{t}+c_1\\ u& =0\end{array}$$

In summary, these are the solution found for $u(t)$

$$\begin{array}{rl}\ln \left(u\right)& =-\frac{2}{t}+c_1\\ u& =0\end{array}$$

For solution $\ln \left(u\right)=-\frac{2}{t}+c_1$, since $u=y^{\prime }\left(t\right)$ then we now have a new first order ode to solve which is

$$\begin{array}{r}\ln \left(y^{\prime }\left(t\right)\right)=-\frac{2}{t}+c_1\end{array}$$

Since the ode has the form $y^{\prime }=f(t)$, then we only need to integrate $f(t)$.

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

For solution $u\left(t\right)=0$, since $u=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=0\end{array}$$

Since the ode has the form $y^{\prime }=f(t)$, then we only need to integrate $f(t)$.

$$\begin{array}{rl}\int dy& =\int 0dt+c_3\\ y& =c_3\end{array}$$

In summary, these are the solution found for $(y)$

$$\begin{array}{rl}y& ={\mathrm{e}}^{-\frac{2}{t}+c_1}t-2{\mathrm{e}}^{c_1}{\mathrm{Ei}}_1\left(\frac{2}{t}\right)+c_2\\ y& =c_3\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_3\\ y& ={\mathrm{e}}^{-\frac{2}{t}+c_1}t-2{\mathrm{e}}^{c_1}{\mathrm{Ei}}_1\left(\frac{2}{t}\right)+c_2\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.196 (sec)

Solve

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

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =t^2\\ \text{(3)}& B& =-2C& =0\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{1+2t}{t^4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =1+2t\\ t& =t^4\end{array}$$

Therefore eq. (4) becomes

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

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 2.17: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-1\\ & =3\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=t^4$. There is a pole at $t=0$ of order $4$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $3$ then the necessary conditions for case one are met. Therefore

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

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

Looking at higher order poles of order $2v$≥$4$ (must be even order for case one).Then for each pole $c$, $[\sqrt{r}{]}_c$ is the sum of terms $\frac{1}{(t-c{)}^i}$ for $2\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ expanded around each pole $c$. Hence

$$\begin{array}{}\text{(1B)}& [\sqrt{r}{]}_c& =\sum _2^v\frac{a_i}{(t-c{)}^i}\end{array}$$

Let $a$ be the coefficient of the term $\frac{1}{(t-c{)}^v}$ in the above where $v$ is the pole order divided by 2. Let $b$ be the coefficient of $\frac{1}{(t-c{)}^{v+1}}$ in $r$ minus the coefficient of $\frac{1}{(t-c{)}^{v+1}}$ in $[\sqrt{r}{]}_c$. Then

$$\begin{array}{rl}{\alpha }_c^{+}& =\frac{1}{2}(\frac{b}{a}+v)\\ {\alpha }_c^{-}& =\frac{1}{2}(-\frac{b}{a}+v)\end{array}$$

The partial fraction decomposition of $r$ is

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

There is pole in $r$ at $t=0$ of order $4$, hence $v=2$. Expanding $\sqrt{r}$ as Laurent series about this pole $c=0$ gives

$$\begin{array}{}\text{(2B)}& [\sqrt{r}{]}_c\approx \frac{1}{t^2}+\frac{1}{t}-\frac{1}{2}+\frac{t}{2}-\frac{5t^2}{8}+\frac{7t^3}{8}+\dots \end{array}$$

Using eq. (1B), taking the sum up to $v=2$ the above becomes

$$\begin{array}{}\text{(3B)}& [\sqrt{r}{]}_c=\frac{1}{t^2}\end{array}$$

The above shows that the coefficient of $\frac{1}{(t-0{)}^2}$ is

$$a=1$$

Now we need to find $b$. let $b$ be the coefficient of the term $\frac{1}{(t-c{)}^{v+1}}$ in $r$ minus the coefficient of the same term but in the sum $[\sqrt{r}{]}_c$ found in eq. (3B). Here $c$ is current pole which is $c=0$. This term becomes $\frac{1}{t^3}$. The coefficient of this term in the sum $[\sqrt{r}{]}_c$ is seen to be $0$ and the coefficient of this term $r$ is found from the partial fraction decomposition from above to be $2$. Therefore

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

Hence

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

Since the order of $r$ at $\mathrm{\infty }$ is $3>2$ 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

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

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$4$	$\frac{1}{t^2}$	$2$	$0$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$3$	$0$	$0$	$1$

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 ${\alpha }_{\mathrm{\infty }}^{+}=0$ then

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

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

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

Substituting the above values in the above results in

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

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(t)$ of degree $d=0$ 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

$$\begin{array}{}\text{(2A)}& p(t)& =1\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{t^2})\left(0\right)+(\left(\frac{2}{t^3}\right)+{(-\frac{1}{t^2})}^2-\left(\frac{1+2t}{t^4}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(t)& =p{e}^{\int \omega dt}\\ & ={\mathrm{e}}^{\int -\frac{1}{t^2}dt}\\ & ={\mathrm{e}}^{\frac{1}{t}}\end{array}$$

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

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

Which simplifies to

$$y_1=1$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-2}{t^2}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{-\frac{2}{t}}}{{\left(y_1\right)}^2}dt\\ & =y_1(t{\mathrm{e}}^{-\frac{2}{t}}-2{\mathrm{Ei}}_1\left(\frac{2}{t}\right))\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_2{y}_1+c_3{y}_2\\ & =c_2\left(1\right)+c_3\left(1(t{\mathrm{e}}^{-\frac{2}{t}}-2{\mathrm{Ei}}_1\left(\frac{2}{t}\right))\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_2+c_3(t{\mathrm{e}}^{-\frac{2}{t}}-2{\mathrm{Ei}}_1\left(\frac{2}{t}\right))\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 25

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

$$y={\mathrm{e}}^{-\frac{2}{t}}{c}_{2}t-2{\mathrm{Ei}}_1\left(\frac{2}{t}\right)c_2+c_1$$

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 
<- LODE missing y successful
 

✓ Mathematica. Time used: 0.017 (sec). Leaf size: 29

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

$$y(t)\to 2c_1\mathrm{ExpIntegralEi}(-\frac{2}{t})+c_1{e}^{-2/t}t+c_2$$

✓ Sympy. Time used: 0.793 (sec). Leaf size: 26

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

$$y\left(t\right)=C_1-C_{2}t{e}^{-\frac{2}{t}}-2C_2\mathrm{Ei}\left(\frac{2e^{i\pi }}{t}\right)$$