4.51 problem 1499

Internal problem ID [9825]
Internal file name [OUTPUT/8768_Monday_June_06_2022_05_25_35_AM_28135746/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1499.
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "unknown"

Maple gives the following as the ode type

[[_3rd_order, _with_linear_symmetries]]

Unable to solve or complete the solution.

\[ \boxed {x^{2} y^{\prime \prime \prime }-\left (x^{2}-2 x \right ) y^{\prime \prime }-\left (x^{2}+\nu ^{2}-\frac {1}{4}\right ) y^{\prime }+\left (x^{2}-2 x +\nu ^{2}-\frac {1}{4}\right ) y=0} \] Unable to solve this ODE.

4.51.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-\left (x^{2}-2 x \right ) \left (\frac {d}{d x}y^{\prime }\right )-\left (x^{2}+\nu ^{2}-\frac {1}{4}\right ) y^{\prime }+\left (x^{2}-2 x +\nu ^{2}-\frac {1}{4}\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\prime \prime } \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=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 -2}{x}, P_{3}\left (x \right )=-\frac {4 \nu ^{2}+4 x^{2}-1}{4 x^{2}}, P_{4}\left (x \right )=\frac {4 \nu ^{2}+4 x^{2}-8 x -1}{4 x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=2 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {1}{4}-\nu ^{2} \\ {} & \circ & x^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{3}\cdot P_{4}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-4 x \left (x -2\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-4 \nu ^{2}-4 x^{2}+1\right ) y^{\prime }+\left (4 \nu ^{2}+4 x^{2}-8 x -1\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) \left (k +r -2\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (2 r -1+2 \nu \right ) \left (2 r -1-2 \nu \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (2 r +1+2 \nu \right ) \left (2 r +1-2 \nu \right )-a_{0} \left (2 r -1+2 \nu \right ) \left (2 r -1-2 \nu \right )\right ) x^{r}+\left (a_{2} \left (2+r \right ) \left (2 r +3+2 \nu \right ) \left (2 r +3-2 \nu \right )-a_{1} \left (2 r +1+2 \nu \right ) \left (2 r +1-2 \nu \right )-4 a_{0} \left (2+r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (2 r +1+2 \nu +2 k \right ) \left (2 r +1-2 \nu +2 k \right )-a_{k} \left (2 r -1+2 \nu +2 k \right ) \left (2 r -1-2 \nu +2 k \right )-4 a_{k -1} \left (k +1+r \right )+4 a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (2 r -1+2 \nu \right ) \left (2 r -1-2 \nu \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}-\nu , \frac {1}{2}+\nu \right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) \left (2 r +1+2 \nu \right ) \left (2 r +1-2 \nu \right )-a_{0} \left (2 r -1+2 \nu \right ) \left (2 r -1-2 \nu \right )=0, a_{2} \left (2+r \right ) \left (2 r +3+2 \nu \right ) \left (2 r +3-2 \nu \right )-a_{1} \left (2 r +1+2 \nu \right ) \left (2 r +1-2 \nu \right )-4 a_{0} \left (2+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (4 \nu ^{2}-4 r^{2}+4 r -1\right )}{4 \nu ^{2} r -4 r^{3}+4 \nu ^{2}-8 r^{2}-5 r -1}, a_{2}=\frac {a_{0} \left (4 \nu ^{2}-8 r^{2}-8 r -9\right )}{4 \nu ^{2} r^{2}-4 r^{4}+12 \nu ^{2} r -24 r^{3}+8 \nu ^{2}-53 r^{2}-51 r -18}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -4 \left (-r -\frac {1}{2}+\nu -k \right ) a_{k +1} \left (r +\frac {1}{2}+\nu +k \right ) \left (k +1+r \right )+a_{k} \left (2 r -1+2 \nu +2 k \right ) \left (-2 r +1+2 \nu -2 k \right )-4 a_{k -1} \left (k +1+r \right )+4 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & -4 \left (-r -\frac {5}{2}+\nu -k \right ) a_{k +3} \left (r +\frac {5}{2}+\nu +k \right ) \left (k +3+r \right )+a_{k +2} \left (2 r +3+2 \nu +2 k \right ) \left (-2 r -3+2 \nu -2 k \right )-4 a_{k +1} \left (k +3+r \right )+4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {4 k^{2} a_{k +2}+8 k r a_{k +2}-4 \nu ^{2} a_{k +2}+4 r^{2} a_{k +2}+4 k a_{k +1}+12 k a_{k +2}+4 r a_{k +1}+12 r a_{k +2}-4 a_{k}+12 a_{k +1}+9 a_{k +2}}{\left (2 r +5-2 \nu +2 k \right ) \left (2 r +5+2 \nu +2 k \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {4 k^{2} a_{k +2}-4 \nu ^{2} a_{k +2}+4 k a_{k +1}+12 k a_{k +2}-4 a_{k}+12 a_{k +1}+9 a_{k +2}}{\left (5-2 \nu +2 k \right ) \left (5+2 \nu +2 k \right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=\frac {4 k^{2} a_{k +2}-4 \nu ^{2} a_{k +2}+4 k a_{k +1}+12 k a_{k +2}-4 a_{k}+12 a_{k +1}+9 a_{k +2}}{\left (5-2 \nu +2 k \right ) \left (5+2 \nu +2 k \right ) \left (k +3\right )}, a_{1}=a_{0}, a_{2}=\frac {a_{0} \left (4 \nu ^{2}-9\right )}{8 \nu ^{2}-18}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\nu \\ {} & {} & a_{k +3}=\frac {4 k^{2} a_{k +2}+8 k \left (\frac {1}{2}-\nu \right ) a_{k +2}-4 \nu ^{2} a_{k +2}+4 \left (\frac {1}{2}-\nu \right )^{2} a_{k +2}+4 k a_{k +1}+12 k a_{k +2}+4 \left (\frac {1}{2}-\nu \right ) a_{k +1}+12 \left (\frac {1}{2}-\nu \right ) a_{k +2}-4 a_{k}+12 a_{k +1}+9 a_{k +2}}{\left (6-4 \nu +2 k \right ) \left (6+2 k \right ) \left (k +\frac {7}{2}-\nu \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\nu }, a_{k +3}=\frac {4 k^{2} a_{k +2}+8 k \left (\frac {1}{2}-\nu \right ) a_{k +2}-4 \nu ^{2} a_{k +2}+4 \left (\frac {1}{2}-\nu \right )^{2} a_{k +2}+4 k a_{k +1}+12 k a_{k +2}+4 \left (\frac {1}{2}-\nu \right ) a_{k +1}+12 \left (\frac {1}{2}-\nu \right ) a_{k +2}-4 a_{k}+12 a_{k +1}+9 a_{k +2}}{\left (6-4 \nu +2 k \right ) \left (6+2 k \right ) \left (k +\frac {7}{2}-\nu \right )}, a_{1}=\frac {a_{0} \left (4 \nu ^{2}-4 \left (\frac {1}{2}-\nu \right )^{2}+1-4 \nu \right )}{4 \nu ^{2} \left (\frac {1}{2}-\nu \right )-4 \left (\frac {1}{2}-\nu \right )^{3}+4 \nu ^{2}-8 \left (\frac {1}{2}-\nu \right )^{2}-\frac {7}{2}+5 \nu }, a_{2}=\frac {a_{0} \left (4 \nu ^{2}-8 \left (\frac {1}{2}-\nu \right )^{2}-13+8 \nu \right )}{4 \nu ^{2} \left (\frac {1}{2}-\nu \right )^{2}-4 \left (\frac {1}{2}-\nu \right )^{4}+12 \nu ^{2} \left (\frac {1}{2}-\nu \right )-24 \left (\frac {1}{2}-\nu \right )^{3}+8 \nu ^{2}-53 \left (\frac {1}{2}-\nu \right )^{2}-\frac {87}{2}+51 \nu }\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}+\nu \\ {} & {} & a_{k +3}=\frac {4 k^{2} a_{k +2}+8 k \left (\frac {1}{2}+\nu \right ) a_{k +2}-4 \nu ^{2} a_{k +2}+4 \left (\frac {1}{2}+\nu \right )^{2} a_{k +2}+4 k a_{k +1}+12 k a_{k +2}+4 \left (\frac {1}{2}+\nu \right ) a_{k +1}+12 \left (\frac {1}{2}+\nu \right ) a_{k +2}-4 a_{k}+12 a_{k +1}+9 a_{k +2}}{\left (6+2 k \right ) \left (6+4 \nu +2 k \right ) \left (k +\frac {7}{2}+\nu \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}+\nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}+\nu }, a_{k +3}=\frac {4 k^{2} a_{k +2}+8 k \left (\frac {1}{2}+\nu \right ) a_{k +2}-4 \nu ^{2} a_{k +2}+4 \left (\frac {1}{2}+\nu \right )^{2} a_{k +2}+4 k a_{k +1}+12 k a_{k +2}+4 \left (\frac {1}{2}+\nu \right ) a_{k +1}+12 \left (\frac {1}{2}+\nu \right ) a_{k +2}-4 a_{k}+12 a_{k +1}+9 a_{k +2}}{\left (6+2 k \right ) \left (6+4 \nu +2 k \right ) \left (k +\frac {7}{2}+\nu \right )}, a_{1}=\frac {a_{0} \left (4 \nu ^{2}-4 \left (\frac {1}{2}+\nu \right )^{2}+1+4 \nu \right )}{4 \nu ^{2} \left (\frac {1}{2}+\nu \right )-4 \left (\frac {1}{2}+\nu \right )^{3}+4 \nu ^{2}-8 \left (\frac {1}{2}+\nu \right )^{2}-\frac {7}{2}-5 \nu }, a_{2}=\frac {a_{0} \left (4 \nu ^{2}-8 \left (\frac {1}{2}+\nu \right )^{2}-13-8 \nu \right )}{4 \nu ^{2} \left (\frac {1}{2}+\nu \right )^{2}-4 \left (\frac {1}{2}+\nu \right )^{4}+12 \nu ^{2} \left (\frac {1}{2}+\nu \right )-24 \left (\frac {1}{2}+\nu \right )^{3}+8 \nu ^{2}-53 \left (\frac {1}{2}+\nu \right )^{2}-\frac {87}{2}-51 \nu }\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}-\nu }\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +\frac {1}{2}+\nu }\right ), a_{k +3}=\frac {4 k^{2} a_{k +2}-4 \nu ^{2} a_{k +2}+4 k a_{k +1}+12 k a_{k +2}-4 a_{k}+12 a_{k +1}+9 a_{k +2}}{\left (5-2 \nu +2 k \right ) \left (5+2 \nu +2 k \right ) \left (k +3\right )}, a_{1}=a_{0}, a_{2}=\frac {a_{0} \left (4 \nu ^{2}-9\right )}{8 \nu ^{2}-18}, b_{k +3}=\frac {4 k^{2} b_{k +2}+8 k \left (\frac {1}{2}-\nu \right ) b_{k +2}-4 \nu ^{2} b_{k +2}+4 \left (\frac {1}{2}-\nu \right )^{2} b_{k +2}+4 k b_{k +1}+12 k b_{k +2}+4 \left (\frac {1}{2}-\nu \right ) b_{k +1}+12 \left (\frac {1}{2}-\nu \right ) b_{k +2}-4 b_{k}+12 b_{k +1}+9 b_{k +2}}{\left (6-4 \nu +2 k \right ) \left (6+2 k \right ) \left (k +\frac {7}{2}-\nu \right )}, b_{1}=\frac {b_{0} \left (4 \nu ^{2}-4 \left (\frac {1}{2}-\nu \right )^{2}+1-4 \nu \right )}{4 \nu ^{2} \left (\frac {1}{2}-\nu \right )-4 \left (\frac {1}{2}-\nu \right )^{3}+4 \nu ^{2}-8 \left (\frac {1}{2}-\nu \right )^{2}-\frac {7}{2}+5 \nu }, b_{2}=\frac {b_{0} \left (4 \nu ^{2}-8 \left (\frac {1}{2}-\nu \right )^{2}-13+8 \nu \right )}{4 \nu ^{2} \left (\frac {1}{2}-\nu \right )^{2}-4 \left (\frac {1}{2}-\nu \right )^{4}+12 \nu ^{2} \left (\frac {1}{2}-\nu \right )-24 \left (\frac {1}{2}-\nu \right )^{3}+8 \nu ^{2}-53 \left (\frac {1}{2}-\nu \right )^{2}-\frac {87}{2}+51 \nu }, c_{k +3}=\frac {4 k^{2} c_{k +2}+8 k \left (\frac {1}{2}+\nu \right ) c_{k +2}-4 \nu ^{2} c_{k +2}+4 \left (\frac {1}{2}+\nu \right )^{2} c_{k +2}+4 k c_{k +1}+12 k c_{k +2}+4 \left (\frac {1}{2}+\nu \right ) c_{k +1}+12 \left (\frac {1}{2}+\nu \right ) c_{k +2}-4 c_{k}+12 c_{k +1}+9 c_{k +2}}{\left (6+2 k \right ) \left (6+4 \nu +2 k \right ) \left (k +\frac {7}{2}+\nu \right )}, c_{1}=\frac {c_{0} \left (4 \nu ^{2}-4 \left (\frac {1}{2}+\nu \right )^{2}+1+4 \nu \right )}{4 \nu ^{2} \left (\frac {1}{2}+\nu \right )-4 \left (\frac {1}{2}+\nu \right )^{3}+4 \nu ^{2}-8 \left (\frac {1}{2}+\nu \right )^{2}-\frac {7}{2}-5 \nu }, c_{2}=\frac {c_{0} \left (4 \nu ^{2}-8 \left (\frac {1}{2}+\nu \right )^{2}-13-8 \nu \right )}{4 \nu ^{2} \left (\frac {1}{2}+\nu \right )^{2}-4 \left (\frac {1}{2}+\nu \right )^{4}+12 \nu ^{2} \left (\frac {1}{2}+\nu \right )-24 \left (\frac {1}{2}+\nu \right )^{3}+8 \nu ^{2}-53 \left (\frac {1}{2}+\nu \right )^{2}-\frac {87}{2}-51 \nu }\right ] \end {array} \]

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying high order exact linear fully integrable 
trying to convert to a linear ODE with constant coefficients 
trying differential order: 3; missing the dependent variable 
Equation is the LCLM of -y(x)+diff(y(x),x), -1/4*(4*nu^2+4*x^2-1)/x^2*y(x)+diff(diff(y(x),x),x) 
trying differential order: 1; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful 
   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 
   <- No Liouvillian solutions exists 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      <- Bessel successful 
   <- special function solution successful 
<- solving the LCLM ode successful `
 

✓ Solution by Maple

Time used: 0.032 (sec). Leaf size: 25

dsolve(x^2*diff(diff(diff(y(x),x),x),x)-(x^2-2*x)*diff(diff(y(x),x),x)-(x^2+nu^2-1/4)*diff(y(x),x)+(x^2-2*x+nu^2-1/4)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = c_{1} {\mathrm e}^{x}+c_{2} \sqrt {x}\, \operatorname {BesselI}\left (\nu , x\right )+c_{3} \sqrt {x}\, \operatorname {BesselK}\left (\nu , x\right ) \]

✓ Solution by Mathematica

Time used: 0.148 (sec). Leaf size: 91

DSolve[(-1/4 + nu^2 - 2*x + x^2)*y[x] - (-1/4 + nu^2 + x^2)*y'[x] - (-2*x + x^2)*y''[x] + x^2*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^x \left (\frac {c_3 x^{\nu +\frac {1}{2}} \operatorname {Gamma}\left (\nu +\frac {1}{2}\right ) \, _1\tilde {F}_1\left (\nu +\frac {1}{2};2 \nu +1;-2 x\right )}{\operatorname {Gamma}\left (\frac {3}{2}-\nu \right )}+c_2 2^{-\nu -\frac {1}{2}} G_{2,3}^{2,1}\left (2 x\left | \begin {array}{c} 1,0 \\ \frac {1}{2}-\nu ,\nu +\frac {1}{2},0 \\ \end {array} \right .\right )+c_1\right ) \]