5.5 problem Problem 1(e)

Internal problem ID [12354]
Internal file name [OUTPUT/11007_Monday_October_02_2023_02_47_52_AM_52665797/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 6. Introduction to Systems of ODEs. Problems page 408
Problem number: Problem 1(e).
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, _linear, _nonhomogeneous]]

Unable to solve or complete the solution.

\[ \boxed {t^{3} y^{\prime \prime }-2 t y^{\prime }+y=t^{4}} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   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 
<- solving first the homogeneous part of the ODE successful`
 

Solution by Maple

Time used: 0.047 (sec). Leaf size: 114

dsolve(t^3*diff(y(t),t$2)-2*t*diff(y(t),t)+y(t)=t^4,y(t), singsol=all)
 

\[ y \left (t \right ) = -\left (\left (-\operatorname {BesselI}\left (0, \frac {1}{t}\right )-\operatorname {BesselI}\left (1, \frac {1}{t}\right )\right ) \left (\int t \left (\operatorname {BesselK}\left (0, \frac {1}{t}\right )-\operatorname {BesselK}\left (1, \frac {1}{t}\right )\right ) {\mathrm e}^{\frac {1}{t}}d t \right )+\left (\int t \left (\operatorname {BesselI}\left (0, \frac {1}{t}\right )+\operatorname {BesselI}\left (1, \frac {1}{t}\right )\right ) {\mathrm e}^{\frac {1}{t}}d t \right ) \left (\operatorname {BesselK}\left (0, \frac {1}{t}\right )-\operatorname {BesselK}\left (1, \frac {1}{t}\right )\right )-\operatorname {BesselK}\left (0, \frac {1}{t}\right ) c_{1} +\operatorname {BesselK}\left (1, \frac {1}{t}\right ) c_{1} -\operatorname {BesselI}\left (0, \frac {1}{t}\right ) c_{2} -\operatorname {BesselI}\left (1, \frac {1}{t}\right ) c_{2} \right ) {\mathrm e}^{-\frac {1}{t}} \]

Solution by Mathematica

Time used: 27.071 (sec). Leaf size: 272

DSolve[t^3*y''[t]-2*t*y'[t]+y[t]==t^4,y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to e^{-1/t} \left (\operatorname {BesselI}\left (0,\frac {1}{t}\right )+\operatorname {BesselI}\left (1,\frac {1}{t}\right )\right ) \left (\int _1^t\frac {2 e^{\frac {2}{K[1]}} \sqrt {\pi } K[1]^3 G_{1,2}^{2,0}\left (\frac {2}{K[1]}| \begin {array}{c} \frac {1}{2} \\ -1,0 \\ \end {array} \right )}{e^{\frac {1}{K[1]}} \sqrt {\pi } \left (\operatorname {BesselI}\left (0,\frac {1}{K[1]}\right )-\operatorname {BesselI}\left (2,\frac {1}{K[1]}\right )\right ) G_{1,2}^{2,0}\left (\frac {2}{K[1]}| \begin {array}{c} \frac {1}{2} \\ -1,0 \\ \end {array} \right )-2 \left (\operatorname {BesselI}\left (0,\frac {1}{K[1]}\right )+\operatorname {BesselI}\left (1,\frac {1}{K[1]}\right )\right ) K_1\left (\frac {1}{K[1]}\right ) K[1]}dK[1]+c_1\right )+G_{1,2}^{2,0}\left (\frac {2}{t}| \begin {array}{c} \frac {1}{2} \\ -1,0 \\ \end {array} \right ) \left (\int _1^t-\frac {2 e^{\frac {1}{K[2]}} \sqrt {\pi } \left (\operatorname {BesselI}\left (0,\frac {1}{K[2]}\right )+\operatorname {BesselI}\left (1,\frac {1}{K[2]}\right )\right ) K[2]^3}{e^{\frac {1}{K[2]}} \sqrt {\pi } \left (\operatorname {BesselI}\left (0,\frac {1}{K[2]}\right )-\operatorname {BesselI}\left (2,\frac {1}{K[2]}\right )\right ) G_{1,2}^{2,0}\left (\frac {2}{K[2]}| \begin {array}{c} \frac {1}{2} \\ -1,0 \\ \end {array} \right )-2 \left (\operatorname {BesselI}\left (0,\frac {1}{K[2]}\right )+\operatorname {BesselI}\left (1,\frac {1}{K[2]}\right )\right ) K_1\left (\frac {1}{K[2]}\right ) K[2]}dK[2]+c_2\right ) \]