1.745 problem 765

Internal problem ID [8883]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 765
Date solved : Monday, October 21, 2024 at 05:23:18 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} \left (x^{2}+2\right ) y^{\prime \prime }+3 x y^{\prime }-y&=0 \end{align*}

1.745.1 Solved as second order ode using Kovacic algorithm

Time used: 0.381 (sec)

Writing the ode as

\begin{align*} \left (x^{2}+2\right ) y^{\prime \prime }+3 x y^{\prime }-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^{2}+2 \\ B &= 3 x\tag {3} \\ C &= -1 \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 {7 x^{2}+20}{4 \left (x^{2}+2\right )^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 7 x^{2}+20\\ t &= 4 \left (x^{2}+2\right )^{2} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= \left ( \frac {7 x^{2}+20}{4 \left (x^{2}+2\right )^{2}}\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 745: 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) \\ &= 4 - 2 \\ &= 2 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 \left (x^{2}+2\right )^{2}\). There is a pole at \(x=i \sqrt {2}\) of order \(2\). There is a pole at \(x=-i \sqrt {2}\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) 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 \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore

\begin{align*} L &= [1, 2, 4, 6, 12] \end{align*}

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

Unable to find solution using case one

Attempting to find a solution using case \(n=2\).

Looking at poles of order 2. The partial fractions decomposition of \(r\) is

\[ r = -\frac {3}{16 \left (x -i \sqrt {2}\right )^{2}}-\frac {3}{16 \left (x +i \sqrt {2}\right )^{2}}-\frac {17 i \sqrt {2}}{32 \left (x -i \sqrt {2}\right )}+\frac {17 i \sqrt {2}}{32 \left (x +i \sqrt {2}\right )} \]

For the pole at \(x=i \sqrt {2}\) let \(b\) be the coefficient of \(\frac {1}{ \left (x -i \sqrt {2}\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {3}{16}}\). Hence

\begin{align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{1, 2, 3\} \end{align*}

For the pole at \(x=-i \sqrt {2}\) let \(b\) be the coefficient of \(\frac {1}{ \left (x +i \sqrt {2}\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {3}{16}}\). Hence

\begin{align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{1, 2, 3\} \end{align*}

Since the order of \(r\) at \(\infty \) is 2 then let \(b\) be the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\) at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\) from

\begin{alignat*}{2} r &= \frac {s}{t} &&= \frac {7 x^{2}+20}{4 \left (x^{2}+2\right )^{2}} \end{alignat*}

Since the \(\text {gcd}(s,t)=1\). This gives \(b={\frac {7}{4}}\). Hence

\begin{align*} E_\infty &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{2\} \end{align*}

The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) for case 2 of Kovacic algorithm.

pole \(c\) location	pole order	\(E_c\)
\(i \sqrt {2}\)	\(2\)	\(\{1, 2, 3\}\)
\(-i \sqrt {2}\)	\(2\)	\(\{1, 2, 3\}\)

Order of \(r\) at \(\infty \)	\(E_\infty \)
\(2\)	\(\{2\}\)

Using the family \(\{e_1,e_2,\dots ,e_\infty \}\) given by

\[ e_1=1,\hspace {3pt} e_2=1,\hspace {3pt} e_\infty =2 \]

Gives a non negative integer \(d\) (the degree of the polynomial \(p(x)\)), which is generated using

\begin{align*} d &= \frac {1}{2} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right )\\ &= \frac {1}{2} \left ( 2 - \left (1+\left (1\right )\right )\right )\\ &= 0 \end{align*}

We now form the following rational function

\begin{align*} \theta &= \frac {1}{2} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {1}{2} \left (\frac {1}{\left (x-\left (i \sqrt {2}\right )\right )}+\frac {1}{\left (x-\left (-i \sqrt {2}\right )\right )}\right ) \\ &= \frac {1}{2 x -2 i \sqrt {2}}+\frac {1}{2 x +2 i \sqrt {2}} \end{align*}

Now we search for a monic polynomial \(p(x)\) of degree \(d=0\) such that

\[ p'''+3 \theta p'' + \left (3 \theta ^2 + 3 \theta ' - 4 r\right )p' + \left (\theta '' + 3 \theta \theta ' + \theta ^3 - 4 r \theta - 2 r' \right ) p = 0 \tag {1A} \]

Since \(d=0\), then letting

\[ p = 1\tag {2A} \]

Substituting \(p\) and \(\theta \) into Eq. (1A) gives

\[ 0 = 0 \]

And solving for \(p\) gives

\[ p = 1 \]

Now that \(p(x)\) is found let

\begin{align*} \phi &= \theta + \frac {p'}{p}\\ &= \frac {1}{2 x -2 i \sqrt {2}}+\frac {1}{2 x +2 i \sqrt {2}} \end{align*}

Let \(\omega \) be the solution of

\begin{align*} \omega ^2 - \phi \omega + \left ( \frac {1}{2} \phi ' + \frac {1}{2} \phi ^2 - r \right ) &= 0 \end{align*}

Substituting the values for \(\phi \) and \(r\) into the above equation gives

\[ w^{2}-\left (\frac {1}{2 x -2 i \sqrt {2}}+\frac {1}{2 x +2 i \sqrt {2}}\right ) w +\frac {7 x^{2}+16}{4 \left (\sqrt {2}+i x \right )^{2} \left (x +i \sqrt {2}\right )^{2}} = 0 \]

Solving for \(\omega \) gives

\begin{align*} \omega &= \frac {x +2 \sqrt {2 x^{2}+4}}{2 x^{2}+4} \end{align*}

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

\begin{align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {x +2 \sqrt {2 x^{2}+4}}{2 x^{2}+4}d x}\\ &= \left (x^{2}+2\right )^{{1}/{4}} {\mathrm e}^{\sqrt {2}\, \operatorname {arcsinh}\left (\frac {\sqrt {2}\, x}{2}\right )} \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 {3 x}{x^{2}+2} \,dx} \\ &= z_1 e^{-\frac {3 \ln \left (x^{2}+2\right )}{4}} \\ &= z_1 \left (\frac {1}{\left (x^{2}+2\right )^{{3}/{4}}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = \frac {{\mathrm e}^{\sqrt {2}\, \operatorname {arcsinh}\left (\frac {\sqrt {2}\, x}{2}\right )}}{\sqrt {x^{2}+2}} \]

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 {3 x}{x^{2}+2} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-\frac {3 \ln \left (x^{2}+2\right )}{2}}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\int \frac {{\mathrm e}^{-2 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {\sqrt {2}\, x}{2}\right )}}{\sqrt {x^{2}+2}}d x\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (\frac {{\mathrm e}^{\sqrt {2}\, \operatorname {arcsinh}\left (\frac {\sqrt {2}\, x}{2}\right )}}{\sqrt {x^{2}+2}}\right ) + c_2 \left (\frac {{\mathrm e}^{\sqrt {2}\, \operatorname {arcsinh}\left (\frac {\sqrt {2}\, x}{2}\right )}}{\sqrt {x^{2}+2}}\left (\int \frac {{\mathrm e}^{-2 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {\sqrt {2}\, x}{2}\right )}}{\sqrt {x^{2}+2}}d x\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

1.745.2 Maple step by step solution

1.745.3 Maple trace

Methods for second order ODEs:
 

1.745.4 Maple dsolve solution

Solving time : 0.027 (sec)
Leaf size : 45

dsolve((x^2+2)*diff(diff(y(x),x),x)+3*x*diff(y(x),x)-y(x) = 0, 
       y(x),singsol=all)
 

\[ y = \frac {c_2 \left (\sqrt {x^{2}+2}+x \right )^{-\sqrt {2}}+c_1 \left (\sqrt {x^{2}+2}+x \right )^{\sqrt {2}}}{\sqrt {x^{2}+2}} \]

1.745.5 Mathematica DSolve solution

Solving time : 0.158 (sec)
Leaf size : 92

DSolve[{(x^2+2)*D[y[x],{x,2}]+3*x*D[y[x],x]-y[x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to \frac {2^{3/4} c_1 \cos \left (2 \sqrt {2} \arcsin \left (\frac {1}{2} \sqrt {2-i \sqrt {2} x}\right )\right )}{\sqrt {\pi } \sqrt {x^2+2}}+\frac {c_2 Q_{-\frac {1}{2}+\sqrt {2}}^{\frac {1}{2}}\left (\frac {i x}{\sqrt {2}}\right )}{\sqrt [4]{x^2+2}} \]