Internal problem ID [1103]
Internal file name [OUTPUT/1104_Sunday_June_05_2022_02_02_39_AM_74431363/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 5 linear second order equations. Section 5.1 Homogeneous linear equations. Page
203
Problem number: 20.
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.
\[ \boxed {\left (3 x -1\right ) y^{\prime \prime }-\left (3 x +2\right ) y^{\prime }+\left (6 x -8\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (3 x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-3 x -2\right ) y^{\prime }+\left (6 x -8\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 }=-\frac {2 \left (3 x -4\right ) y}{3 x -1}+\frac {\left (3 x +2\right ) y^{\prime }}{3 x -1} \\ \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 (3 x +2\right ) y^{\prime }}{3 x -1}+\frac {2 \left (3 x -4\right ) y}{3 x -1}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=\frac {1}{3}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {3 x +2}{3 x -1}, P_{3}\left (x \right )=\frac {2 \left (3 x -4\right )}{3 x -1}\right ] \\ {} & \circ & \left (x -\frac {1}{3}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{3} \\ {} & {} & \left (\left (x -\frac {1}{3}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{3}}}}=-1 \\ {} & \circ & \left (x -\frac {1}{3}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{3} \\ {} & {} & \left (\left (x -\frac {1}{3}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{3}}}}=0 \\ {} & \circ & x =\frac {1}{3}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=\frac {1}{3}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=\frac {1}{3} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (3 x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-3 x -2\right ) y^{\prime }+\left (6 x -8\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {1}{3}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & 3 u \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-3 u -3\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (6 u -6\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 3 a_{0} r \left (-2+r \right ) u^{-1+r}+\left (3 a_{1} \left (1+r \right ) \left (-1+r \right )-3 a_{0} \left (2+r \right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (3 a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )-3 a_{k} \left (k +r +2\right )+6 a_{k -1}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 3 r \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 3 a_{1} \left (1+r \right ) \left (-1+r \right )-3 a_{0} \left (2+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )+a_{k} \left (-3 k -3 r -6\right )+6 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 3 a_{k +2} \left (k +r +2\right ) \left (k +r \right )+a_{k +1} \left (-3 k -9-3 r \right )+6 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {k a_{k +1}+r a_{k +1}-2 a_{k}+3 a_{k +1}}{\left (k +r +2\right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {k a_{k +1}-2 a_{k}+3 a_{k +1}}{\left (k +2\right ) k} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =0\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =0 \\ {} & {} & a_{k +2}=\frac {k a_{k +1}-2 a_{k}+3 a_{k +1}}{\left (k +2\right ) k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=\frac {k a_{k +1}-2 a_{k}+5 a_{k +1}}{\left (k +4\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +2}, a_{k +2}=\frac {k a_{k +1}-2 a_{k}+5 a_{k +1}}{\left (k +4\right ) \left (k +2\right )}, 9 a_{1}-12 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{3}\right )^{k +2}, a_{k +2}=\frac {k a_{k +1}-2 a_{k}+5 a_{k +1}}{\left (k +4\right ) \left (k +2\right )}, 9 a_{1}-12 a_{0}=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 -> 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: 0.031 (sec). Leaf size: 177
dsolve((3*x-1)*diff(y(x),x$2)-(3*x+2)*diff(y(x),x)+(6*x-8)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {6615 \,{\mathrm e}^{-\frac {x \left (i \sqrt {7}-1\right )}{2}} \left (x -\frac {1}{3}\right )^{2} \left (\left (\left (-\frac {23 x}{35}+\frac {296}{735}\right ) \sqrt {7}+i x -\frac {248 i}{105}\right ) c_{1} \operatorname {KummerM}\left (\frac {1}{2}-\frac {5 i \sqrt {7}}{14}, 3, \frac {i \sqrt {7}\, \left (3 x -1\right )}{3}\right )-\left (\left (-\frac {x}{5}-\frac {8}{105}\right ) \sqrt {7}+i x -\frac {152 i}{105}\right ) c_{2} \operatorname {KummerU}\left (\frac {1}{2}-\frac {5 i \sqrt {7}}{14}, 3, \frac {i \sqrt {7}\, \left (3 x -1\right )}{3}\right )+2 \left (i+\frac {3 \sqrt {7}}{49}\right ) c_{1} \operatorname {KummerM}\left (-\frac {1}{2}-\frac {5 i \sqrt {7}}{14}, 3, \frac {i \sqrt {7}\, \left (3 x -1\right )}{3}\right )+\frac {\left (i+\frac {5 \sqrt {7}}{7}\right ) \operatorname {KummerU}\left (-\frac {1}{2}-\frac {5 i \sqrt {7}}{14}, 3, \frac {i \sqrt {7}\, \left (3 x -1\right )}{3}\right ) c_{2}}{5}\right )}{-225 \sqrt {7}+21 i} \]
✓ Solution by Mathematica
Time used: 0.13 (sec). Leaf size: 109
DSolve[(3*x-1)*y''[x]-(3*x+2)*y'[x]+(6*x-8)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to 4 e^{\frac {1}{6} \left (1-i \sqrt {7}\right ) (3 x-1)} (1-3 x)^2 \left (c_1 \operatorname {HypergeometricU}\left (\frac {3}{2}-\frac {5 i}{2 \sqrt {7}},3,\frac {1}{3} i \sqrt {7} (3 x-1)\right )+c_2 L_{-\frac {3}{2}+\frac {5 i}{2 \sqrt {7}}}^2\left (\frac {1}{3} i \sqrt {7} (3 x-1)\right )\right ) \]