Internal problem ID [9529]
Internal file name [OUTPUT/8469_Monday_June_06_2022_03_09_23_AM_11369270/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1200.
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.
\[ \boxed {x^{2} y^{\prime \prime }+2 y^{\prime } x^{2}-v \left (v -1\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+2 y^{\prime } x^{2}+\left (-v^{2}+v \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 }=-2 y^{\prime }+\frac {v \left (v -1\right ) y}{x^{2}} \\ \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 }+2 y^{\prime }-\frac {v \left (v -1\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 )=2, P_{3}\left (x \right )=-\frac {v \left (v -1\right )}{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}}}=0 \\ {} & \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}}}=-v \left (v -1\right ) \\ {} & \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 )+2 y^{\prime } x^{2}-v \left (v -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^{2}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1+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 -1+r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (v -1+r \right ) \left (-v +r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (v -1+r +k \right ) \left (-v +r +k \right )+2 a_{k -1} \left (k -1+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (v -1+r \right ) \left (-v +r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{v , -v +1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (v -1+r +k \right ) \left (-v +r +k \right )+2 a_{k -1} \left (k -1+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (v +r +k \right ) \left (-v +r +k +1\right )+2 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k +r \right )}{\left (v +r +k \right ) \left (-v +r +k +1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =v \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k +v \right )}{\left (2 v +k \right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =v \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +v}, a_{k +1}=-\frac {2 a_{k} \left (k +v \right )}{\left (2 v +k \right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-v +1 \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k -v +1\right )}{\left (k +1\right ) \left (-2 v +2+k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-v +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -v +1}, a_{k +1}=-\frac {2 a_{k} \left (k -v +1\right )}{\left (k +1\right ) \left (-2 v +2+k \right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +v}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -v +1}\right ), a_{k +1}=-\frac {2 a_{k} \left (k +v \right )}{\left (2 v +k \right ) \left (k +1\right )}, b_{k +1}=-\frac {2 b_{k} \left (k -v +1\right )}{\left (k +1\right ) \left (-2 v +2+k \right )}\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: 0.016 (sec). Leaf size: 27
dsolve(x^2*diff(diff(y(x),x),x)+2*x^2*diff(y(x),x)-v*(v-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \sqrt {x}\, {\mathrm e}^{-x} \left (c_{1} \operatorname {BesselI}\left (v -\frac {1}{2}, x\right )+c_{2} \operatorname {BesselK}\left (v -\frac {1}{2}, x\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.039 (sec). Leaf size: 45
DSolve[(1 - v)*v*y[x] + 2*x^2*y'[x] + x^2*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x} \sqrt {x} \left (c_1 \operatorname {BesselJ}\left (v-\frac {1}{2},-i x\right )+c_2 \operatorname {BesselY}\left (v-\frac {1}{2},-i x\right )\right ) \]