Internal problem ID [819]
Internal file name [OUTPUT/819_Sunday_June_05_2022_01_50_31_AM_27651476/index.tex]
Book: Elementary differential equations and boundary value problems, 11th ed., Boyce, DiPrima,
Meade
Section: Chapter 4.1, Higher order linear differential equations. General theory. page
173
Problem number: 17.
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.
\[ \boxed {t y^{\prime \prime \prime }+2 y^{\prime \prime }-y^{\prime }+y t=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t \left (\frac {d}{d t}y^{\prime \prime }\right )+2 \frac {d}{d t}y^{\prime }-y^{\prime }+y t =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d t}y^{\prime \prime } \\ \square & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (t \right )=\frac {2}{t}, P_{3}\left (t \right )=-\frac {1}{t}, P_{4}\left (t \right )=1\right ] \\ {} & \circ & t \cdot P_{2}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t \cdot P_{2}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=2 \\ {} & \circ & t^{2}\cdot P_{3}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t^{2}\cdot P_{3}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=0 \\ {} & \circ & t^{3}\cdot P_{4}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t^{3}\cdot P_{4}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=0 \\ {} & \circ & t =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t \cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & t \cdot y=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) t^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d t}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d t}y^{\prime }=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +r +1\right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t \cdot \left (\frac {d}{d t}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot \left (\frac {d}{d t}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 ) t^{k +r -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & t \cdot \left (\frac {d}{d t}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +r +1\right ) \left (k +r \right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r^{2} \left (-1+r \right ) t^{-2+r}+\left (a_{1} \left (1+r \right )^{2} r -a_{0} r \right ) t^{-1+r}+\left (a_{2} \left (2+r \right )^{2} \left (1+r \right )-a_{1} \left (1+r \right )\right ) t^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +2} \left (k +2+r \right )^{2} \left (k +r +1\right )-a_{k +1} \left (k +r +1\right )+a_{k -1}\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2} \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} t \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right )^{2} r -a_{0} r =0, a_{2} \left (2+r \right )^{2} \left (1+r \right )-a_{1} \left (1+r \right )=0\right ] \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +2} \left (k +2+r \right )^{2} \left (k +r +1\right )+\left (-k -r -1\right ) a_{k +1}+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +3} \left (k +3+r \right )^{2} \left (k +2+r \right )+\left (-k -2-r \right ) a_{k +2}+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k a_{k +2}+r a_{k +2}-a_{k}+2 a_{k +2}}{\left (k +3+r \right )^{2} \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {k a_{k +2}-a_{k}+2 a_{k +2}}{\left (k +3\right )^{2} \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +3}=\frac {k a_{k +2}-a_{k}+2 a_{k +2}}{\left (k +3\right )^{2} \left (k +2\right )}, 0=0, 4 a_{2}-a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +3}=\frac {k a_{k +2}-a_{k}+3 a_{k +2}}{\left (k +4\right )^{2} \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +1}, a_{k +3}=\frac {k a_{k +2}-a_{k}+3 a_{k +2}}{\left (k +4\right )^{2} \left (k +3\right )}, 4 a_{1}-a_{0}=0, 18 a_{2}-2 a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{k +1}\right ), a_{k +3}=\frac {k a_{k +2}-a_{k}+2 a_{k +2}}{\left (k +3\right )^{2} \left (k +2\right )}, 0=0, 4 a_{2}-a_{1}=0, b_{k +3}=\frac {k b_{k +2}-b_{k}+3 b_{k +2}}{\left (k +4\right )^{2} \left (k +3\right )}, 4 b_{1}-b_{0}=0, 18 b_{2}-2 b_{1}=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 trying Louvillian solutions for 3rd order ODEs, imprimitive case -> pFq: Equivalence to the 3F2 or one of its 3 confluent cases under a power @ Moebius trying a solution in terms of MeijerG functions -> pFq: Equivalence to the 3F2 or one of its 3 confluent cases under a power @ Moebius trying a solution in terms of MeijerG functions checking if the LODE is of Euler type Calling dsolve with: (t-1)/t*y(t)-(t-1)/t*diff(y(t),t)+diff(diff(y(t),t),t) = 0 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 <- differential factorization successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 159
dsolve(t*diff(y(t),t$3)+2*diff(y(t),t$2)-diff(y(t),t)+t*y(t)=0,y(t), singsol=all)
\[ y \left (t \right ) = {\mathrm e}^{-\frac {t \left (i \sqrt {3}-1\right )}{2}} \left (\operatorname {KummerM}\left (\frac {1}{2}-\frac {i \sqrt {3}}{6}, 1, i \sqrt {3}\, t \right ) \left (\int \operatorname {KummerU}\left (\frac {1}{2}-\frac {i \sqrt {3}}{6}, 1, i \sqrt {3}\, t \right ) {\mathrm e}^{-\frac {t \left (i \sqrt {3}+3\right )}{2}}d t \right ) c_{3} -\operatorname {KummerU}\left (\frac {1}{2}-\frac {i \sqrt {3}}{6}, 1, i \sqrt {3}\, t \right ) \left (\int \operatorname {KummerM}\left (\frac {1}{2}-\frac {i \sqrt {3}}{6}, 1, i \sqrt {3}\, t \right ) {\mathrm e}^{-\frac {t \left (i \sqrt {3}+3\right )}{2}}d t \right ) c_{3} +c_{1} \operatorname {KummerM}\left (\frac {1}{2}-\frac {i \sqrt {3}}{6}, 1, i \sqrt {3}\, t \right )+c_{2} \operatorname {KummerU}\left (\frac {1}{2}-\frac {i \sqrt {3}}{6}, 1, i \sqrt {3}\, t \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.639 (sec). Leaf size: 520
DSolve[t*y'''[t]+2*y''[t]-y'[t]+t*y[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{\frac {1}{2} \left (t-i \sqrt {3} t\right )} \left (c_3 \operatorname {HypergeometricU}\left (\frac {1}{6} \left (3-i \sqrt {3}\right ),1,i \sqrt {3} t\right ) \int _1^t\frac {2 e^{\frac {1}{2} i \left (3 i+\sqrt {3}\right ) K[1]} \operatorname {LaguerreL}\left (\frac {1}{6} i \left (3 i+\sqrt {3}\right ),i \sqrt {3} K[1]\right )}{\left (-1-i \sqrt {3}\right ) K[1] \left (\operatorname {Hypergeometric1F1}\left (\frac {1}{6} \left (9-i \sqrt {3}\right ),2,i \sqrt {3} K[1]\right ) \operatorname {HypergeometricU}\left (\frac {1}{6} \left (3-i \sqrt {3}\right ),1,i \sqrt {3} K[1]\right )+\operatorname {HypergeometricU}\left (\frac {1}{6} \left (9-i \sqrt {3}\right ),2,i \sqrt {3} K[1]\right ) \operatorname {LaguerreL}\left (\frac {1}{6} i \left (3 i+\sqrt {3}\right ),i \sqrt {3} K[1]\right )\right )}dK[1]+c_3 \operatorname {LaguerreL}\left (\frac {1}{6} i \left (3 i+\sqrt {3}\right ),i \sqrt {3} t\right ) \int _1^t-\frac {2 i e^{\frac {1}{2} i \left (3 i+\sqrt {3}\right ) K[2]} \operatorname {HypergeometricU}\left (\frac {1}{6} \left (3-i \sqrt {3}\right ),1,i \sqrt {3} K[2]\right )}{\left (-i+\sqrt {3}\right ) K[2] \left (\operatorname {Hypergeometric1F1}\left (\frac {1}{6} \left (9-i \sqrt {3}\right ),2,i \sqrt {3} K[2]\right ) \operatorname {HypergeometricU}\left (\frac {1}{6} \left (3-i \sqrt {3}\right ),1,i \sqrt {3} K[2]\right )+\operatorname {HypergeometricU}\left (\frac {1}{6} \left (9-i \sqrt {3}\right ),2,i \sqrt {3} K[2]\right ) \operatorname {LaguerreL}\left (\frac {1}{6} i \left (3 i+\sqrt {3}\right ),i \sqrt {3} K[2]\right )\right )}dK[2]+c_1 \operatorname {HypergeometricU}\left (\frac {1}{6} \left (3-i \sqrt {3}\right ),1,i \sqrt {3} t\right )+c_2 \operatorname {LaguerreL}\left (\frac {1}{6} i \left (3 i+\sqrt {3}\right ),i \sqrt {3} t\right )\right ) \]