Internal problem ID [9542]
Internal file name [OUTPUT/8482_Monday_June_06_2022_03_11_38_AM_10692059/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1213.
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.
\[ \boxed {x^{2} y^{\prime \prime }+\left (x^{3}+1\right ) x y^{\prime }-y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (x^{4}+x \right ) y^{\prime }-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 {y}{x^{2}}-\frac {\left (x^{3}+1\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 (x^{3}+1\right ) y^{\prime }}{x}-\frac {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 {x^{3}+1}{x}, P_{3}\left (x \right )=-\frac {1}{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}}}=1 \\ {} & \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}}}=-1 \\ {} & \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 )+\left (x^{3}+1\right ) x y^{\prime }-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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..4 \\ {} & {} & 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 (1+r \right ) \left (-1+r \right ) x^{r}+a_{1} \left (2+r \right ) r \,x^{1+r}+a_{2} \left (3+r \right ) \left (1+r \right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k} \left (k +r +1\right ) \left (k +r -1\right )+a_{k -3} \left (k -3+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 1\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (2+r \right ) r =0, a_{2} \left (3+r \right ) \left (1+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r +1\right ) \left (k +r -1\right )+a_{k -3} \left (k -3+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +3} \left (k +4+r \right ) \left (k +2+r \right )+a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {a_{k} \left (k +r \right )}{\left (k +4+r \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +3}=-\frac {a_{k} \left (k -1\right )}{\left (k +3\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +3}=-\frac {a_{k} \left (k -1\right )}{\left (k +3\right ) \left (k +1\right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +3}=-\frac {a_{k} \left (k +1\right )}{\left (k +5\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +3}=-\frac {a_{k} \left (k +1\right )}{\left (k +5\right ) \left (k +3\right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1}\right ), a_{k +3}=-\frac {a_{k} \left (k -1\right )}{\left (k +3\right ) \left (k +1\right )}, a_{1}=0, a_{2}=0, b_{k +3}=-\frac {b_{k} \left (k +1\right )}{\left (k +5\right ) \left (k +3\right )}, b_{1}=0, b_{2}=0\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 successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 52
dsolve(x^2*diff(diff(y(x),x),x)+(x^3+1)*x*diff(y(x),x)-y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {x^{3}}{6}} x^{\frac {3}{2}} \left (c_{1} \operatorname {BesselI}\left (-\frac {1}{6}, \frac {x^{3}}{6}\right )+c_{1} \operatorname {BesselI}\left (\frac {5}{6}, \frac {x^{3}}{6}\right )+c_{2} \left (\operatorname {BesselK}\left (\frac {1}{6}, \frac {x^{3}}{6}\right )-\operatorname {BesselK}\left (\frac {5}{6}, \frac {x^{3}}{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.143 (sec). Leaf size: 54
DSolve[-y[x] + x*(1 + x^3)*y'[x] + x^2*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\sqrt [3]{3} c_1 \operatorname {Hypergeometric1F1}\left (-\frac {1}{3},\frac {1}{3},-\frac {x^3}{3}\right )}{x}+\frac {c_2 x \operatorname {Hypergeometric1F1}\left (\frac {1}{3},\frac {5}{3},-\frac {x^3}{3}\right )}{\sqrt [3]{3}} \]