1.723 problem 740

Internal problem ID [8861]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 740
Date solved : Monday, October 21, 2024 at 05:22:57 PM
CAS classification : [_Hermite]

Solve

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

1.723.1 Solved as second order ode using Kovacic algorithm

Time used: 0.250 (sec)

Writing the ode as

\begin{align*} 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 &= 1 \\ 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^{2}-10}{4}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= x^{2}-10\\ t &= 4 \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= \left ( \frac {x^{2}}{4}-\frac {5}{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 723: 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) \\ &= 0 - 2 \\ &= -2 \end{align*}

There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. 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. Therefore

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

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

Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = -2\) then

\begin{alignat*}{3} v &= \frac {-O_r(\infty )}{2} &&= \frac {2}{2} &&= 1 \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}^{1} a_i x^i \tag {8} \end{align*}

Let \(a\) be the coefficient of \(x^v=x^1\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is

\[ \sqrt r \approx \frac {x}{2}-\frac {5}{2 x}-\frac {25}{4 x^{3}}-\frac {125}{4 x^{5}}-\frac {3125}{16 x^{7}}-\frac {21875}{16 x^{9}}-\frac {328125}{32 x^{11}}-\frac {2578125}{32 x^{13}} + \dots \tag {9} \]

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

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

From Eq. (9) the sum up to \(v=1\) gives

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

Now we need to find \(b\), where \(b\) be the coefficient of \(x^{v-1} = x^{0}=1\) 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

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

This shows that the coefficient of \(1\) in the above is \(0\). Now we need to find the coefficient of \(1\) in \(r\). How this is done depends on if \(v=0\) or not. Since \(v=1\) 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 \(1\) in \(r\) will be the coefficient this term in the quotient. Doing long division gives

\begin{align*} r &= \frac {s}{t} \\ &= \frac {x^{2}-10}{4} \\ &= Q + \frac {R}{4} \\ &= \left (\frac {x^{2}}{4}-\frac {5}{2}\right ) + \left ( 0\right ) \\ &= \frac {x^{2}}{4}-\frac {5}{2} \end{align*}

We see that the coefficient of the term \(\frac {1}{x}\) in the quotient is \(-{\frac {5}{2}}\). Now \(b\) can be found.

\begin{align*} b &= \left (-{\frac {5}{2}}\right )-\left (0\right )\\ &= -{\frac {5}{2}} \end{align*}

Hence

\begin{alignat*}{3} [\sqrt r]_\infty &= \frac {x}{2}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {-{\frac {5}{2}}}{{\frac {1}{2}}} - 1 \right ) &&= -3\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {-{\frac {5}{2}}}{{\frac {1}{2}}} - 1 \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

\[ r=\frac {x^{2}}{4}-\frac {5}{2} \]

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(-2\)	\(\frac {x}{2}\)	\(-3\)	\(2\)

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 \(\alpha _\infty ^{-} = 2\), and since there are no poles then

\begin{align*} d &= \alpha _\infty ^{-} \\ &= 2 \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*}

The above gives

\begin{align*} \omega &= (-) [\sqrt r]_\infty \\ &= 0 + (-) \left ( \frac {x}{2} \right ) \\ &= -\frac {x}{2}\\ &= -\frac {x}{2} \end{align*}

Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=2\) 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^{2}+a_{1} x +a_{0}\tag {2A} \end{align*}

Substituting the above in eq. (1A) gives

\begin{align*} \left (2\right ) + 2 \left (-\frac {x}{2}\right ) \left (2 x +a_{1}\right ) + \left ( \left (-{\frac {1}{2}}\right ) + \left (-\frac {x}{2}\right )^2 - \left (\frac {x^{2}}{4}-\frac {5}{2}\right ) \right ) &= 0\\ a_{1} x +2 a_{0}+2 = 0 \end{align*}

Solving for the coefficients \(a_i\) in the above using method of undetermined coefficients gives

\[ \{a_{0} = -1, a_{1} = 0\} \]

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

\begin{align*} p(x) &= x^{2}-1 \end{align*}

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

\begin{align*} z_1(x) &= p e^{ \int \omega \,dx}\\ & = \left (x^{2}-1\right ) {\mathrm e}^{\int -\frac {x}{2}d x}\\ & = \left (x^{2}-1\right ) {\mathrm e}^{-\frac {x^{2}}{4}}\\ & = \left (x^{2}-1\right ) {\mathrm e}^{-\frac {x^{2}}{4}} \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}{1} \,dx} \\ &= z_1 e^{\frac {x^{2}}{4}} \\ &= z_1 \left ({\mathrm e}^{\frac {x^{2}}{4}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = x^{2}-1 \]

The second solution \(y_2\) to the original ode is found using reduction of order

\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \]

Substituting gives

\begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-x}{1} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{\frac {x^{2}}{2}}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\int \frac {{\mathrm e}^{\frac {x^{2}}{2}}}{\left (x^{2}-1\right )^{2}}d x\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (x^{2}-1\right ) + c_2 \left (x^{2}-1\left (\int \frac {{\mathrm e}^{\frac {x^{2}}{2}}}{\left (x^{2}-1\right )^{2}}d x\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

1.723.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }-x y^{\prime }+2 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k} \\ \square & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} k \,x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) x^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) x^{k} \\ & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & {} & \moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )-a_{k} \left (k -2\right )\right ) x^{k}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+3 k +2\right ) a_{k +2}-a_{k} \left (k -2\right )=0 \\ \bullet & {} & \textrm {Recursion relation; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +2}=\frac {a_{k} \left (k -2\right )}{k^{2}+3 k +2} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{2}=-a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=A_{2} x^{2}+A_{1} x -a_{0} \end {array} \]

1.723.3 Maple trace

Methods for second order ODEs:
 

1.723.4 Maple dsolve solution

Solving time : 0.012 (sec)
Leaf size : 39

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

\[ y = -2 \,{\mathrm e}^{\frac {x^{2}}{2}} c_1 x +\left (x -1\right ) \left (x +1\right ) \left (\sqrt {\pi }\, \sqrt {2}\, \operatorname {erfi}\left (\frac {\sqrt {2}\, x}{2}\right ) c_1 +c_2 \right ) \]

1.723.5 Mathematica DSolve solution

Solving time : 0.136 (sec)
Leaf size : 54

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

\[ y(x)\to \frac {1}{4} c_2 \left (\sqrt {2 \pi } \left (x^2-1\right ) \text {erfi}\left (\frac {x}{\sqrt {2}}\right )-2 e^{\frac {x^2}{2}} x\right )+c_1 \left (x^2-1\right ) \]