Internal problem ID [9526]
Internal file name [OUTPUT/8466_Monday_June_06_2022_03_08_48_AM_27091423/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1197.
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.
\[ \boxed {x^{2} y^{\prime \prime }-\left (x^{2}-2 x \right ) y^{\prime }-\left (x +a \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (-x^{2}+2 x \right ) y^{\prime }+\left (-a -x \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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\left (x +a \right ) y}{x^{2}}+\frac {\left (-2+x \right ) y^{\prime }}{x} \\ \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 {\left (-2+x \right ) y^{\prime }}{x}-\frac {\left (x +a \right ) y}{x^{2}}=0 \\ \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 {-2+x}{x}, P_{3}\left (x \right )=-\frac {x +a}{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}}}=-a \\ {} & \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} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )-x \left (-2+x \right ) y^{\prime }+\left (-a -x \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..1 \\ {} & {} & 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 =1..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^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} \left (-r^{2}+a -r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (-k^{2}-2 k r -r^{2}+a -k -r \right )-a_{k -1} \left (k +r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}-a +r =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}, -\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (2 r +1\right ) k +r^{2}-a +r \right ) a_{k}-a_{k -1} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+\left (2 r +1\right ) \left (k +1\right )+r^{2}-a +r \right ) a_{k +1}-a_{k} \left (k +r +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +r +1\right )}{-k^{2}-2 k r -r^{2}+a -3 k -3 r -2} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2}-\frac {\sqrt {1+4 a}}{2} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}\right )}{-k^{2}-2 k \left (-\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}\right )-\left (-\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}\right )^{2}+a -3 k -\frac {1}{2}+\frac {3 \sqrt {1+4 a}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2}-\frac {\sqrt {1+4 a}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}}, a_{k +1}=-\frac {a_{k} \left (k +\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}\right )}{-k^{2}-2 k \left (-\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}\right )-\left (-\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}\right )^{2}+a -3 k -\frac {1}{2}+\frac {3 \sqrt {1+4 a}}{2}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2}+\frac {\sqrt {1+4 a}}{2} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right )}{-k^{2}-2 k \left (-\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right )-\left (-\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right )^{2}+a -3 k -\frac {1}{2}-\frac {3 \sqrt {1+4 a}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2}+\frac {\sqrt {1+4 a}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}}, a_{k +1}=-\frac {a_{k} \left (k +\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right )}{-k^{2}-2 k \left (-\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right )-\left (-\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right )^{2}+a -3 k -\frac {1}{2}-\frac {3 \sqrt {1+4 a}}{2}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k -\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}}\right ), b_{k +1}=-\frac {b_{k} \left (k +\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}\right )}{-k^{2}-2 k \left (-\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}\right )-\left (-\frac {1}{2}-\frac {\sqrt {1+4 a}}{2}\right )^{2}+a -3 k -\frac {1}{2}+\frac {3 \sqrt {1+4 a}}{2}}, c_{k +1}=-\frac {c_{k} \left (k +\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right )}{-k^{2}-2 k \left (-\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right )-\left (-\frac {1}{2}+\frac {\sqrt {1+4 a}}{2}\right )^{2}+a -3 k -\frac {1}{2}-\frac {3 \sqrt {1+4 a}}{2}}\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 <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 43
dsolve(x^2*diff(diff(y(x),x),x)-(x^2-2*x)*diff(y(x),x)-(x+a)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{\frac {x}{2}} \left (\operatorname {BesselK}\left (\frac {\sqrt {4 a +1}}{2}, \frac {x}{2}\right ) c_{2} +\operatorname {BesselI}\left (\frac {\sqrt {4 a +1}}{2}, \frac {x}{2}\right ) c_{1} \right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.042 (sec). Leaf size: 67
DSolve[(-a - x)*y[x] - (-2*x + x^2)*y'[x] + x^2*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {e^{x/2} \left (c_1 \operatorname {BesselJ}\left (\frac {1}{2} \sqrt {4 a+1},-\frac {i x}{2}\right )+c_2 \operatorname {BesselY}\left (\frac {1}{2} \sqrt {4 a+1},-\frac {i x}{2}\right )\right )}{\sqrt {x}} \]