1.427 problem 439
Internal
problem
ID
[8565]
Book
:
Collection
of
Kovacic
problems
Section
:
section
1
Problem
number
:
439
Date
solved
:
Monday, October 21, 2024 at 05:09:46 PM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} \left (-x^{2}+4\right ) y^{\prime \prime }+x y^{\prime }+2 y&=0 \end{align*}
1.427.1 Solved as second order ode using Kovacic algorithm
Time used: 1.022 (sec)
Writing the ode as
\begin{align*} \left (-x^{2}+4\right ) 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^{2}+4 \\ 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 {11 x^{2}-24}{4 \left (x^{2}-4\right )^{2}}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= 11 x^{2}-24\\ t &= 4 \left (x^{2}-4\right )^{2} \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(x) &= \left ( \frac {11 x^{2}-24}{4 \left (x^{2}-4\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 427: 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 \(t=4 \left (x^{2}-4\right )^{2}\).
There is a pole at \(x=2\) of order \(2\). There is a pole at \(x=-2\) 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
\[
r = \frac {17}{32 \left (x -2\right )}-\frac {17}{32 \left (x +2\right )}+\frac {5}{16 \left (x +2\right )^{2}}+\frac {5}{16 \left (x -2\right )^{2}}
\]
For the pole at \(x=2\) let \(b\)
be the coefficient of \(\frac {1}{ \left (x -2\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {5}{16}}\). Hence
\begin{align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{-1, 2, 5\} \end{align*}
For the pole at \(x=-2\) let \(b\) be the coefficient of \(\frac {1}{ \left (x +2\right )^{2}}\) in the partial fractions decomposition of \(r\) given
above. Therefore \(b={\frac {5}{16}}\). Hence
\begin{align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{-1, 2, 5\} \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 {11 x^{2}-24}{4 \left (x^{2}-4\right )^{2}} \end{alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b={\frac {11}{4}}\). 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 |
\(E_c\) |
| | |
\(2\) | \(2\) | \(\{-1, 2, 5\}\) |
| | |
\(-2\) | \(2\) | \(\{-1, 2, 5\}\) |
| | |
| |
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 )\\ &= 2 \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 (2\right )\right )}+\frac {-1}{\left (x-\left (-2\right )\right )}\right ) \\ &= -\frac {1}{2 \left (x -2\right )}-\frac {1}{2 \left (x +2\right )} \end{align*}
Now we search for a monic polynomial \(p(x)\) of degree \(d=2\) 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 \(d=2\), then letting
\[ p = x^{2}+a_{1} x +a_{0}\tag {2A} \]
Substituting \(p\) and \(\theta \) into Eq. (1A) gives
\[
\frac {11 x^{2} a_{1}+16 \left (6+a_{0}\right ) x +36 a_{1}}{\left (x^{2}-4\right )^{2}} = 0
\]
And solving for \(p\) gives
\[ p = x^{2}-6 \]
Now that \(p(x)\) is found let
\begin{align*} \phi &= \theta + \frac {p'}{p}\\ &= \frac {2 x}{x^{2}-6}-\frac {1}{2 \left (x -2\right )}-\frac {1}{2 \left (x +2\right )} \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
\[
w^{2}-\left (\frac {2 x}{x^{2}-6}-\frac {1}{2 \left (x -2\right )}-\frac {1}{2 \left (x +2\right )}\right ) w +\frac {-11 x^{4}+74 x^{2}-128}{4 x^{6}-56 x^{4}+256 x^{2}-384} = 0
\]
Solving for \(\omega \) gives
\begin{align*} \omega &= \frac {2 \sqrt {3}\, x^{2} \sqrt {x^{2}-4}+x^{3}-8 \sqrt {3}\, \sqrt {x^{2}-4}-2 x}{2 \left (x^{2}-6\right ) \left (x -2\right ) \left (x +2\right )} \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 {2 \sqrt {3}\, x^{2} \sqrt {x^{2}-4}+x^{3}-8 \sqrt {3}\, \sqrt {x^{2}-4}-2 x}{2 \left (x^{2}-6\right ) \left (x -2\right ) \left (x +2\right )}d x}\\ &= \frac {\sqrt {x^{2}-6}\, \left (x +\sqrt {x^{2}-4}\right )^{\sqrt {3}} {\mathrm e}^{-\frac {\operatorname {arctanh}\left (\frac {\left (\sqrt {2}\, \sqrt {3}\, x -4\right ) \sqrt {2}}{2 \sqrt {x^{2}-4}}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {\left (4+\sqrt {2}\, \sqrt {3}\, x \right ) \sqrt {2}}{2 \sqrt {x^{2}-4}}\right )}{2}}}{\left (x +2\right )^{{1}/{4}} \left (x -2\right )^{{1}/{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}{-x^{2}+4} \,dx} \\
&= z_1 e^{\frac {\ln \left (x^{2}-4\right )}{4}} \\
&= z_1 \left (\left (x^{2}-4\right )^{{1}/{4}}\right ) \\
\end{align*}
Which simplifies to
\[
y_1 = \sqrt {x^{2}-6}\, \left (x +\sqrt {x^{2}-4}\right )^{\sqrt {3}} {\mathrm e}^{-\frac {\operatorname {arctanh}\left (\frac {x \sqrt {6}-4}{\sqrt {2 x^{2}-8}}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {4+x \sqrt {6}}{\sqrt {2 x^{2}-8}}\right )}{2}}
\]
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 {x}{-x^{2}+4} \,dx}}{\left (y_1\right )^2} \,dx \\
&= y_1 \int \frac { e^{\frac {\ln \left (x^{2}-4\right )}{2}}}{\left (y_1\right )^2} \,dx \\
&= y_1 \left (\int \frac {\sqrt {x^{2}-4}\, \left (x +\sqrt {x^{2}-4}\right )^{-2 \sqrt {3}} {\mathrm e}^{\operatorname {arctanh}\left (\frac {x \sqrt {6}-4}{\sqrt {2 x^{2}-8}}\right )+\operatorname {arctanh}\left (\frac {4+x \sqrt {6}}{\sqrt {2 x^{2}-8}}\right )}}{x^{2}-6}d x\right ) \\
\end{align*}
Therefore
the solution is
\begin{align*}
y &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left (\sqrt {x^{2}-6}\, \left (x +\sqrt {x^{2}-4}\right )^{\sqrt {3}} {\mathrm e}^{-\frac {\operatorname {arctanh}\left (\frac {x \sqrt {6}-4}{\sqrt {2 x^{2}-8}}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {4+x \sqrt {6}}{\sqrt {2 x^{2}-8}}\right )}{2}}\right ) + c_2 \left (\sqrt {x^{2}-6}\, \left (x +\sqrt {x^{2}-4}\right )^{\sqrt {3}} {\mathrm e}^{-\frac {\operatorname {arctanh}\left (\frac {x \sqrt {6}-4}{\sqrt {2 x^{2}-8}}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {4+x \sqrt {6}}{\sqrt {2 x^{2}-8}}\right )}{2}}\left (\int \frac {\sqrt {x^{2}-4}\, \left (x +\sqrt {x^{2}-4}\right )^{-2 \sqrt {3}} {\mathrm e}^{\operatorname {arctanh}\left (\frac {x \sqrt {6}-4}{\sqrt {2 x^{2}-8}}\right )+\operatorname {arctanh}\left (\frac {4+x \sqrt {6}}{\sqrt {2 x^{2}-8}}\right )}}{x^{2}-6}d x\right )\right ) \\
\end{align*}
Will add steps showing solving for IC soon.
1.427.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-x^{2}+4\right ) \left (\frac {d}{d x}y^{\prime }\right )+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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {2 y}{x^{2}-4}+\frac {x y^{\prime }}{x^{2}-4} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {x y^{\prime }}{x^{2}-4}-\frac {2 y}{x^{2}-4}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {x}{x^{2}-4}, P_{3}\left (x \right )=-\frac {2}{x^{2}-4}\right ] \\ {} & \circ & \left (x +2\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=-\frac {1}{2} \\ {} & \circ & \left (x +2\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=0 \\ {} & \circ & x =-2\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-2 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x^{2}-4\right ) \left (\frac {d}{d x}y^{\prime }\right )-x y^{\prime }-2 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -2\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-4 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-u +2\right ) \left (\frac {d}{d u}y \left (u \right )\right )-2 y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a_{0} r \left (-3+2 r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right ) \left (2 k -1+2 r \right )+a_{k} \left (k^{2}+2 k r +r^{2}-2 k -2 r -2\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r \left (-3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {3}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -4 \left (k +1+r \right ) \left (k -\frac {1}{2}+r \right ) a_{k +1}+\left (k^{2}+\left (2 r -2\right ) k +r^{2}-2 r -2\right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (k^{2}+2 k r +r^{2}-2 k -2 r -2\right ) a_{k}}{2 \left (k +1+r \right ) \left (2 k -1+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {\left (k^{2}-2 k -2\right ) a_{k}}{2 \left (k +1\right ) \left (2 k -1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=\frac {\left (k^{2}-2 k -2\right ) a_{k}}{2 \left (k +1\right ) \left (2 k -1\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}, a_{k +1}=\frac {\left (k^{2}-2 k -2\right ) a_{k}}{2 \left (k +1\right ) \left (2 k -1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & a_{k +1}=\frac {\left (k^{2}+k -\frac {11}{4}\right ) a_{k}}{2 \left (k +\frac {5}{2}\right ) \left (2 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {3}{2}}, a_{k +1}=\frac {\left (k^{2}+k -\frac {11}{4}\right ) a_{k}}{2 \left (k +\frac {5}{2}\right ) \left (2 k +2\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k +\frac {3}{2}}, a_{k +1}=\frac {\left (k^{2}+k -\frac {11}{4}\right ) a_{k}}{2 \left (k +\frac {5}{2}\right ) \left (2 k +2\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +2\right )^{k +\frac {3}{2}}\right ), a_{k +1}=\frac {\left (k^{2}-2 k -2\right ) a_{k}}{2 \left (k +1\right ) \left (2 k -1\right )}, b_{k +1}=\frac {\left (k^{2}+k -\frac {11}{4}\right ) b_{k}}{2 \left (k +\frac {5}{2}\right ) \left (2 k +2\right )}\right ] \end {array} \]
1.427.3 Maple trace
Methods for second order ODEs:
1.427.4 Maple dsolve solution
Solving time : 0.008
(sec)
Leaf size : 37
dsolve((-x^2+4)*diff(diff(y(x),x),x)+x*diff(y(x),x)+2*y(x) = 0,
y(x),singsol=all)
\[
y = \left (x^{2}-4\right )^{{3}/{4}} \left (\operatorname {LegendreP}\left (-\frac {1}{2}+\sqrt {3}, \frac {3}{2}, \frac {x}{2}\right ) c_1 +\operatorname {LegendreQ}\left (-\frac {1}{2}+\sqrt {3}, \frac {3}{2}, \frac {x}{2}\right ) c_2 \right )
\]
1.427.5 Mathematica DSolve solution
Solving time : 0.076
(sec)
Leaf size : 58
DSolve[{(4-x^2)*D[y[x],{x,2}]+x*D[y[x],x]+2*y[x]==0,{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to \left (x^2-4\right )^{3/4} \left (c_1 P_{-\frac {1}{2}+\sqrt {3}}^{\frac {3}{2}}\left (\frac {x}{2}\right )+c_2 Q_{-\frac {1}{2}+\sqrt {3}}^{\frac {3}{2}}\left (\frac {x}{2}\right )\right )
\]