2.1.727 problem 744

Internal problem ID [9311]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 744
Date solved : Sunday, November 10, 2024 at 04:36:15 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} x y^{\prime \prime }+x y^{\prime }-2 y&=0 \end{align*}

Solved as second order ode using Kovacic algorithm

Time used: 0.254 (sec)

Writing the ode as

\begin{align*} x y^{\prime \prime }+x y^{\prime }-2 y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= x \\ B &= x\tag {3} \\ C &= -2 \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end{align*}

Then (2) becomes

\begin{align*} z''(x) = r z(x)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives

\begin{align*} r &= \frac {x +8}{4 x}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= x +8\\ t &= 4 x \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= \left ( \frac {x +8}{4 x}\right ) z(x)\tag {7} \end{align*}

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

\begin{align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end{align*}

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 \(\infty \). The following table summarizes these cases.

Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 2.727: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 1 - 1 \\ &= 0 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 x\). There is a pole at \(x=0\) of order \(1\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(0\) then the necessary conditions for case one are met. Therefore

\begin{align*} L &= [1] \end{align*}

Attempting to find a solution using case \(n=1\).

Looking at poles of order 1. For the pole at \(x = 0\) of order 1 then

\begin{align*} [\sqrt r]_c &= 0 \\ \alpha _c^+ &= 1 \\ \alpha _c^- &= 1 \end{align*}

Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = 0\) then

\begin{alignat*}{3} v &= \frac {-O_r(\infty )}{2} &&= \frac {0}{2} &&= 0 \end{alignat*}

\([\sqrt r]_\infty \) is the sum of terms involving \(x^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore

\begin{align*} [\sqrt r]_\infty &= \sum _{i=0}^{v} a_i x^i \\ &= \sum _{i=0}^{0} a_i x^i \tag {8} \end{align*}

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

\[ \sqrt r \approx \frac {1}{2}+\frac {2}{x}-\frac {4}{x^{2}}+\frac {16}{x^{3}}-\frac {80}{x^{4}}+\frac {448}{x^{5}}-\frac {2688}{x^{6}}+\frac {16896}{x^{7}} + \dots \tag {9} \]

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

\[ a = {\frac {1}{2}} \]

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

\begin{align*} [\sqrt r]_\infty &= \sum _{i=0}^{0} a_i x^i \\ &= {\frac {1}{2}} \tag {10} \end{align*}

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 \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence

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

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{align*} r &= \frac {s}{t} \\ &= \frac {x +8}{4 x} \\ &= Q + \frac {R}{4 x}\\ &= \left ({\frac {1}{4}}\right ) + \left ( \frac {2}{x}\right ) \\ &= \frac {1}{4}+\frac {2}{x} \end{align*}

Since the degree of \(t\) is \(1\), then we see that the coefficient of the term \(1\) in the remainder \(R\) is \(8\). Dividing this by leading coefficient in \(t\) which is \(4\) gives \(2\). Now \(b\) can be found.

\begin{align*} b &= \left (2\right )-\left (0\right )\\ &= 2 \end{align*}

Hence

\begin{alignat*}{3} [\sqrt r]_\infty &= {\frac {1}{2}}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {2}{{\frac {1}{2}}} - 0 \right ) &&= 2\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {2}{{\frac {1}{2}}} - 0 \right ) &&= -2 \end{alignat*}

The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is

\[ r=\frac {x +8}{4 x} \]

pole \(c\) location	pole order	\([\sqrt r]_c\)	\(\alpha _c^{+}\)	\(\alpha _c^{-}\)
\(0\)	\(1\)	\(0\)	\(0\)	\(1\)

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

Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using

\begin{align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end{align*}

Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{+} = 2\) then

\begin{align*} d &= \alpha _\infty ^{+} - \left ( \alpha _{c_1}^{-} \right ) \\ &= 2 - \left ( 1 \right ) \\ &= 1 \end{align*}

Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using

\begin{align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}

Substituting the above values in the above results in

\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (+) [\sqrt r]_\infty \\ &= \frac {1}{x} + \left ( {\frac {1}{2}} \right ) \\ &= \frac {1}{x}+\frac {1}{2}\\ &= \frac {1}{x}+\frac {1}{2} \end{align*}

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{align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end{align*}

Let

\begin{align*} p(x) &= x +a_{0}\tag {2A} \end{align*}

Substituting the above in eq. (1A) gives

\begin{align*} \left (0\right ) + 2 \left (\frac {1}{x}+\frac {1}{2}\right ) \left (1\right ) + \left ( \left (-\frac {1}{x^{2}}\right ) + \left (\frac {1}{x}+\frac {1}{2}\right )^2 - \left (\frac {x +8}{4 x}\right ) \right ) &= 0\\ \frac {2-a_{0}}{x} = 0 \end{align*}

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{align*} p(x) &= 2+x \end{align*}

Therefore the first solution to the ode \(z'' = r z\) is

\begin{align*} z_1(x) &= p e^{ \int \omega \,dx}\\ & = \left (2+x\right ) {\mathrm e}^{\int \left (\frac {1}{x}+\frac {1}{2}\right )d x}\\ & = \left (2+x\right ) {\mathrm e}^{\frac {x}{2}+\ln \left (x \right )}\\ & = \left (2+x \right ) x \,{\mathrm e}^{\frac {x}{2}} \end{align*}

The first solution to the original ode in \(y\) is found from

\begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {x}{x} \,dx} \\ &= z_1 e^{-\frac {x}{2}} \\ &= z_1 \left ({\mathrm e}^{-\frac {x}{2}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = \left (2+x \right ) x \]

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{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {x}{x} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-x}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (-\frac {{\mathrm e}^{-x}}{4 x}+\frac {\operatorname {Ei}_{1}\left (x \right )}{2}+\frac {{\mathrm e}^{-x}}{-8-4 x}\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (\left (2+x \right ) x\right ) + c_2 \left (\left (2+x \right ) x\left (-\frac {{\mathrm e}^{-x}}{4 x}+\frac {\operatorname {Ei}_{1}\left (x \right )}{2}+\frac {{\mathrm e}^{-x}}{-8-4 x}\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Maple step by step solution

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 dsolve solution

Solving time : 0.007 (sec)
Leaf size : 28

dsolve(x*diff(diff(y(x),x),x)+diff(y(x),x)*x-2*y(x) = 0, 
       y(x),singsol=all)
 

\[ y = -\frac {\left (x +1\right ) c_{2} {\mathrm e}^{-x}}{2}+\left (c_{1} +\frac {\operatorname {Ei}_{1}\left (x \right ) c_{2}}{2}\right ) x \left (x +2\right ) \]

Mathematica DSolve solution

Solving time : 0.087 (sec)
Leaf size : 39

DSolve[{x*D[y[x],{x,2}]+x*D[y[x],x]-2*y[x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to c_1 x (x+2)-\frac {1}{2} c_2 e^{-x} \left (e^x (x+2) x \operatorname {ExpIntegralEi}(-x)+x+1\right ) \]