Internal problem ID [820]
Internal file name [OUTPUT/820_Sunday_June_05_2022_01_50_33_AM_70836635/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: 20.
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.
\[ \boxed {\left (2-t \right ) y^{\prime \prime \prime }+\left (2 t -3\right ) y^{\prime \prime }-y^{\prime } t +y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2-t \right ) \left (\frac {d}{d t}y^{\prime \prime }\right )+\left (2 t -3\right ) \left (\frac {d}{d t}y^{\prime }\right )-y^{\prime } t +y=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}=2\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 -3}{-2+t}, P_{3}\left (t \right )=\frac {t}{-2+t}, P_{4}\left (t \right )=-\frac {1}{-2+t}\right ] \\ {} & \circ & \left (-2+t \right )\cdot P_{2}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =2 \\ {} & {} & \left (\left (-2+t \right )\cdot P_{2}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}2}}}=-1 \\ {} & \circ & \left (-2+t \right )^{2}\cdot P_{3}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =2 \\ {} & {} & \left (\left (-2+t \right )^{2}\cdot P_{3}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}2}}}=0 \\ {} & \circ & \left (-2+t \right )^{3}\cdot P_{4}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =2 \\ {} & {} & \left (\left (-2+t \right )^{3}\cdot P_{4}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}2}}}=0 \\ {} & \circ & t =2\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=2\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=2 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (-2+t \right ) \left (\frac {d}{d t}y^{\prime \prime }\right )+\left (-2 t +3\right ) \left (\frac {d}{d t}y^{\prime }\right )+y^{\prime } t -y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} t =u +2\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & u \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-2 u -1\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (u +2\right ) \left (\frac {d}{d u}y \left (u \right )\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 \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^{m}\cdot \left (\frac {d}{d u}\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}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u \cdot \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) \left (k +r -2\right ) u^{k +r -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+r \right ) \left (-3+r \right ) u^{-2+r}+\left (a_{1} \left (1+r \right ) r \left (-2+r \right )-2 a_{0} r \left (-2+r \right )\right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r -1\right )-2 a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )+a_{k} \left (k +r -1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+r \right ) \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, 3\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -1\right ) \left (a_{k +2} \left (k +2+r \right ) \left (k +1+r \right )-2 a_{k +1} k -2 a_{k +1} r +a_{k}-2 a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 a_{k +1} k +2 a_{k +1} r -a_{k}+2 a_{k +1}}{\left (k +2+r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {2 a_{k +1} k -a_{k}+2 a_{k +1}}{\left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=\frac {2 a_{k +1} k -a_{k}+2 a_{k +1}}{\left (k +2\right ) \left (k +1\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =-2+t \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (-2+t \right )^{k}, a_{k +2}=\frac {2 a_{k +1} k -a_{k}+2 a_{k +1}}{\left (k +2\right ) \left (k +1\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {2 a_{k +1} k -a_{k}+4 a_{k +1}}{\left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +1}, a_{k +2}=\frac {2 a_{k +1} k -a_{k}+4 a_{k +1}}{\left (k +3\right ) \left (k +2\right )}, -2 a_{1}+2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =-2+t \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (-2+t \right )^{k +1}, a_{k +2}=\frac {2 a_{k +1} k -a_{k}+4 a_{k +1}}{\left (k +3\right ) \left (k +2\right )}, -2 a_{1}+2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +2}=\frac {2 a_{k +1} k -a_{k}+8 a_{k +1}}{\left (k +5\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =3 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +3}, a_{k +2}=\frac {2 a_{k +1} k -a_{k}+8 a_{k +1}}{\left (k +5\right ) \left (k +4\right )}, 12 a_{1}-6 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =-2+t \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (-2+t \right )^{k +3}, a_{k +2}=\frac {2 a_{k +1} k -a_{k}+8 a_{k +1}}{\left (k +5\right ) \left (k +4\right )}, 12 a_{1}-6 a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (-2+t \right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (-2+t \right )^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} \left (-2+t \right )^{k +3}\right ), a_{k +2}=\frac {2 k a_{k +1}-a_{k}+2 a_{k +1}}{\left (k +2\right ) \left (k +1\right )}, 0=0, b_{k +2}=\frac {2 k b_{k +1}-b_{k}+4 b_{k +1}}{\left (k +3\right ) \left (k +2\right )}, -2 b_{1}+2 b_{0}=0, c_{k +2}=\frac {2 k c_{k +1}-c_{k}+8 c_{k +1}}{\left (k +5\right ) \left (k +4\right )}, 12 c_{1}-6 c_{0}=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 LCLM of -1/t*y(t)+diff(y(t),t), -y(t)+diff(y(t),t), (-1-1/t)*y(t)+diff(y(t),t) trying differential order: 1; missing the dependent variable checking if the LODE is of Euler type <- LODE of Euler type successful Euler equation successful trying differential order: 1; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful trying differential order: 1; missing the dependent variable checking if the LODE is of Euler type checking if the LODE is of Euler type exponential solutions successful <- differential factorization successful <- solving the LCLM ode successful `
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 16
dsolve([(2-t)*diff(y(t),t$3)+(2*t-3)*diff(y(t),t$2)-t*diff(y(t),t)+y(t)=0,exp(t)],singsol=all)
\[ y \left (t \right ) = {\mathrm e}^{t} \left (c_{3} t +c_{2} \right )+c_{1} t \]
✓ Solution by Mathematica
Time used: 0.079 (sec). Leaf size: 28
DSolve[(2-t)*y'''[t]+(2*t-3)*y''[t]-t*y'[t]+y[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to t \left (c_2 e^t+c_1\right )+(c_3-4 c_2) e^t \]