Internal problem ID [9380]
Internal file name [OUTPUT/8317_Monday_June_06_2022_02_42_36_AM_28248887/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1048.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime \prime }-4 x y^{\prime }+\left (3 x^{2}+2 n -1\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }-4 x y^{\prime }+\left (3 x^{2}+2 n -1\right ) 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 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 =\max \left (0, -m \right )}{\sum }}a_{k} x^{k +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )+m}{\sum }}a_{k -m} x^{k} \\ {} & \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 ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{2}-a_{0} \left (1-2 n \right )+\left (6 a_{3}-a_{1} \left (5-2 n \right )\right ) x +\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )-a_{k} \left (4 k -2 n +1\right )+3 a_{k -2}\right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [2 a_{2}-a_{0} \left (1-2 n \right )=0, 6 a_{3}-a_{1} \left (5-2 n \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{2}=-a_{0} n +\frac {1}{2} a_{0}, a_{3}=-\frac {1}{3} a_{1} n +\frac {5}{6} a_{1}\right \} \\ \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}+\left (-4 k +2 n -1\right ) a_{k}+3 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+3 k +8\right ) a_{k +4}+\left (-4 k -9+2 n \right ) a_{k +2}+3 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=\frac {4 k a_{k +2}-2 n a_{k +2}-3 a_{k}+9 a_{k +2}}{k^{2}+7 k +12}, a_{2}=-a_{0} n +\frac {1}{2} a_{0}, a_{3}=-\frac {1}{3} a_{1} n +\frac {5}{6} a_{1}\right ] \end {array} \]
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 <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.079 (sec). Leaf size: 37
dsolve(diff(diff(y(x),x),x)-4*x*diff(y(x),x)+(3*x^2+2*n-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x \,{\mathrm e}^{\frac {x^{2}}{2}} \left (\operatorname {KummerU}\left (-\frac {n}{2}+\frac {1}{2}, \frac {3}{2}, x^{2}\right ) c_{2} +\operatorname {KummerM}\left (-\frac {n}{2}+\frac {1}{2}, \frac {3}{2}, x^{2}\right ) c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.031 (sec). Leaf size: 37
DSolve[(-1 + 2*n + 3*x^2)*y[x] - 4*x*y'[x] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{\frac {x^2}{2}} \left (c_1 \operatorname {HermiteH}(n,x)+c_2 \operatorname {Hypergeometric1F1}\left (-\frac {n}{2},\frac {1}{2},x^2\right )\right ) \]