Internal problem ID [9818]
Internal file name [OUTPUT/8761_Monday_June_06_2022_05_24_46_AM_47336188/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1492.
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.
\[ \boxed {x^{2} y^{\prime \prime \prime }-3 \left (x -m \right ) x y^{\prime \prime }+\left (2 x^{2}+4 \left (n -m \right ) x +m \left (2 m -1\right )\right ) y^{\prime }-2 n \left (2 x -2 m +1\right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-3 \left (x -m \right ) x \left (\frac {d}{d x}y^{\prime }\right )+\left (2 x^{2}+4 \left (n -m \right ) x +m \left (2 m -1\right )\right ) y^{\prime }-2 n \left (2 x -2 m +1\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 {3 \left (-x +m \right )}{x}, P_{3}\left (x \right )=\frac {2 m^{2}-4 x m +4 x n +2 x^{2}-m}{x^{2}}, P_{4}\left (x \right )=\frac {2 n \left (-2 x +2 m -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}}}=3 m \\ {} & \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}}}=2 m^{2}-m \\ {} & \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} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )+3 \left (-x +m \right ) x \left (\frac {d}{d x}y^{\prime }\right )+\left (2 m^{2}-4 x m +4 x n +2 x^{2}-m \right ) y^{\prime }+2 n \left (-2 x +2 m -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..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 =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 (r -2+2 m \right ) \left (r +m -1\right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (r -1+2 m \right ) \left (r +m \right )-a_{0} \left (-4 n m +4 m r -4 n r +3 r^{2}+2 n -3 r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (r -1+2 m +k \right ) \left (r +m +k \right )-a_{k} \left (3 k^{2}+4 m k -4 n k +6 k r -4 n m +4 m r -4 n r +3 r^{2}-3 k +2 n -3 r \right )+2 a_{k -1} \left (k -1+r -2 n \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (r -2+2 m \right ) \left (r +m -1\right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1-m , 2-2 m \right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (r -1+2 m \right ) \left (r +m \right )-a_{0} \left (-4 n m +4 m r -4 n r +3 r^{2}+2 n -3 r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (r -1+2 m +k \right ) \left (r +m +k \right )-3 k^{2} a_{k}+\left (-6 a_{k} r -4 a_{k} m +\left (4 n +3\right ) a_{k}+2 a_{k -1}\right ) k -3 r^{2} a_{k}+\left (-4 a_{k} m +\left (4 n +3\right ) a_{k}+2 a_{k -1}\right ) r +4 n a_{k} m -2 n a_{k}-4 a_{k -1} n -2 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (r +2 m +k \right ) \left (r +m +k +1\right )-3 \left (k +1\right )^{2} a_{k +1}+\left (-6 r a_{k +1}-4 m a_{k +1}+\left (4 n +3\right ) a_{k +1}+2 a_{k}\right ) \left (k +1\right )-3 r^{2} a_{k +1}+\left (-4 m a_{k +1}+\left (4 n +3\right ) a_{k +1}+2 a_{k}\right ) r +4 n a_{k +1} m -2 n a_{k +1}-4 n a_{k}-2 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {3 k^{2} a_{k +1}+4 k m a_{k +1}-4 k n a_{k +1}+6 k r a_{k +1}-4 n a_{k +1} m +4 m r a_{k +1}-4 n r a_{k +1}+3 r^{2} a_{k +1}-2 a_{k} k +3 k a_{k +1}+4 m a_{k +1}+4 n a_{k}-2 n a_{k +1}-2 a_{k} r +3 r a_{k +1}}{\left (k +2+r \right ) \left (r +2 m +k \right ) \left (r +m +k +1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {3 k^{2} a_{k +1}+4 k m a_{k +1}-4 k n a_{k +1}-4 n a_{k +1} m -2 a_{k} k +3 k a_{k +1}+4 m a_{k +1}+4 n a_{k}-2 n a_{k +1}}{\left (k +2\right ) \left (2 m +k \right ) \left (1+m +k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {3 k^{2} a_{k +1}+4 k m a_{k +1}-4 k n a_{k +1}-4 n a_{k +1} m -2 a_{k} k +3 k a_{k +1}+4 m a_{k +1}+4 n a_{k}-2 n a_{k +1}}{\left (k +2\right ) \left (2 m +k \right ) \left (1+m +k \right )}, a_{1} \left (2 m -1\right ) m -a_{0} \left (-4 n m +2 n \right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1-m \\ {} & {} & a_{k +2}=\frac {3 k^{2} a_{k +1}+4 k m a_{k +1}-4 k n a_{k +1}+6 k \left (1-m \right ) a_{k +1}-4 n a_{k +1} m +4 m \left (1-m \right ) a_{k +1}-4 n \left (1-m \right ) a_{k +1}+3 \left (1-m \right )^{2} a_{k +1}-2 a_{k} k +3 k a_{k +1}+4 m a_{k +1}+4 n a_{k}-2 n a_{k +1}-2 a_{k} \left (1-m \right )+3 \left (1-m \right ) a_{k +1}}{\left (k +3-m \right ) \left (1+m +k \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1-m \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1-m}, a_{k +2}=\frac {3 k^{2} a_{k +1}+4 k m a_{k +1}-4 k n a_{k +1}+6 k \left (1-m \right ) a_{k +1}-4 n a_{k +1} m +4 m \left (1-m \right ) a_{k +1}-4 n \left (1-m \right ) a_{k +1}+3 \left (1-m \right )^{2} a_{k +1}-2 a_{k} k +3 k a_{k +1}+4 m a_{k +1}+4 n a_{k}-2 n a_{k +1}-2 a_{k} \left (1-m \right )+3 \left (1-m \right ) a_{k +1}}{\left (k +3-m \right ) \left (1+m +k \right ) \left (k +2\right )}, a_{1} \left (2-m \right ) m -a_{0} \left (-4 n m +4 m \left (1-m \right )-4 n \left (1-m \right )+3 \left (1-m \right )^{2}+2 n -3+3 m \right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2-2 m \\ {} & {} & a_{k +2}=\frac {3 k^{2} a_{k +1}+4 k m a_{k +1}-4 k n a_{k +1}+6 k \left (2-2 m \right ) a_{k +1}-4 n a_{k +1} m +4 m \left (2-2 m \right ) a_{k +1}-4 n \left (2-2 m \right ) a_{k +1}+3 \left (2-2 m \right )^{2} a_{k +1}-2 a_{k} k +3 k a_{k +1}+4 m a_{k +1}+4 n a_{k}-2 n a_{k +1}-2 a_{k} \left (2-2 m \right )+3 \left (2-2 m \right ) a_{k +1}}{\left (k +4-2 m \right ) \left (k +2\right ) \left (k +3-m \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2-2 m \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2-2 m}, a_{k +2}=\frac {3 k^{2} a_{k +1}+4 k m a_{k +1}-4 k n a_{k +1}+6 k \left (2-2 m \right ) a_{k +1}-4 n a_{k +1} m +4 m \left (2-2 m \right ) a_{k +1}-4 n \left (2-2 m \right ) a_{k +1}+3 \left (2-2 m \right )^{2} a_{k +1}-2 a_{k} k +3 k a_{k +1}+4 m a_{k +1}+4 n a_{k}-2 n a_{k +1}-2 a_{k} \left (2-2 m \right )+3 \left (2-2 m \right ) a_{k +1}}{\left (k +4-2 m \right ) \left (k +2\right ) \left (k +3-m \right )}, a_{1} \left (3-2 m \right ) \left (2-m \right )-a_{0} \left (-4 n m +4 m \left (2-2 m \right )-4 n \left (2-2 m \right )+3 \left (2-2 m \right )^{2}+2 n -6+6 m \right )=0\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 +1-m}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +2-2 m}\right ), a_{k +2}=\frac {3 k^{2} a_{k +1}+4 k m a_{k +1}-4 k n a_{k +1}-4 m n a_{k +1}-2 k a_{k}+3 k a_{k +1}+4 m a_{k +1}+4 n a_{k}-2 n a_{k +1}}{\left (k +2\right ) \left (2 m +k \right ) \left (1+m +k \right )}, a_{1} \left (2 m -1\right ) m -a_{0} \left (-4 n m +2 n \right )=0, b_{k +2}=\frac {3 k^{2} b_{k +1}+4 k m b_{k +1}-4 k n b_{k +1}+6 k \left (1-m \right ) b_{k +1}-4 n b_{k +1} m +4 m \left (1-m \right ) b_{k +1}-4 n \left (1-m \right ) b_{k +1}+3 \left (1-m \right )^{2} b_{k +1}-2 b_{k} k +3 k b_{k +1}+4 m b_{k +1}+4 n b_{k}-2 n b_{k +1}-2 b_{k} \left (1-m \right )+3 \left (1-m \right ) b_{k +1}}{\left (k +3-m \right ) \left (1+m +k \right ) \left (k +2\right )}, b_{1} \left (2-m \right ) m -b_{0} \left (-4 n m +4 m \left (1-m \right )-4 n \left (1-m \right )+3 \left (1-m \right )^{2}+2 n -3+3 m \right )=0, c_{k +2}=\frac {3 k^{2} c_{k +1}+4 k m c_{k +1}-4 k n c_{k +1}+6 k \left (2-2 m \right ) c_{k +1}-4 n c_{k +1} m +4 m \left (2-2 m \right ) c_{k +1}-4 n \left (2-2 m \right ) c_{k +1}+3 \left (2-2 m \right )^{2} c_{k +1}-2 c_{k} k +3 k c_{k +1}+4 m c_{k +1}+4 n c_{k}-2 n c_{k +1}-2 c_{k} \left (2-2 m \right )+3 \left (2-2 m \right ) c_{k +1}}{\left (k +4-2 m \right ) \left (k +2\right ) \left (k +3-m \right )}, c_{1} \left (3-2 m \right ) \left (2-m \right )-c_{0} \left (-4 n m +4 m \left (2-2 m \right )-4 n \left (2-2 m \right )+3 \left (2-2 m \right )^{2}+2 n -6+6 m \right )=0\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 2nd symmetric power of diff(diff(y(x),x),x)+1/x*(-x+m)*diff(y(x),x)+n/x*y(x) = 0 -> Attempting now to solve this lower order ODE 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: 0.063 (sec). Leaf size: 39
dsolve(x^2*diff(diff(diff(y(x),x),x),x)-3*(x-m)*x*diff(diff(y(x),x),x)+(2*x^2+4*(n-m)*x+m*(2*m-1))*diff(y(x),x)-2*n*(2*x-2*m+1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {KummerM}\left (-n , m , x\right )^{2}+c_{2} \operatorname {KummerU}\left (-n , m , x\right )^{2}+c_{3} \operatorname {KummerM}\left (-n , m , x\right ) \operatorname {KummerU}\left (-n , m , x\right ) \]
✓ Solution by Mathematica
Time used: 0.19 (sec). Leaf size: 43
DSolve[-2*n*(1 - 2*m + 2*x)*y[x] + (m*(-1 + 2*m) + 4*(-m + n)*x + 2*x^2)*y'[x] - 3*x*(-m + x)*y''[x] + x^2*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 \operatorname {HypergeometricU}(-n,m,x) L_n^{m-1}(x)+c_1 \operatorname {HypergeometricU}(-n,m,x)^2+c_3 L_n^{m-1}(x){}^2 \]