problem 1485

Internal problem ID [9811]
Internal ﬁle name [OUTPUT/8754_Monday_June_06_2022_05_23_59_AM_93084041/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1485.
ODE order: 3.
ODE degree: 1.

\[ \boxed {\left (x -2\right ) x y^{\prime \prime \prime }-\left (x -2\right ) x y^{\prime \prime }-2 y^{\prime }+2 y=0} \] Unable to solve this ODE.

4.37.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x -2\right ) x \left (\frac {d}{d x}y^{\prime \prime }\right )-\left (x -2\right ) x \left (\frac {d}{d x}y^{\prime }\right )-2 y^{\prime }+2 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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-1, P_{3}\left (x \right )=-\frac {2}{\left (x -2\right ) x}, P_{4}\left (x \right )=\frac {2}{\left (x -2\right ) x}\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}}}=0 \\ {} & \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}}}=0 \\ {} & \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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=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} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{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 +1+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^{m}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\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 -3+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-3+m}{\sum }}a_{k +3-m} \left (k +3-m +r \right ) \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a_{0} r \left (-1+r \right ) \left (-2+r \right ) x^{-2+r}+\left (-2 a_{1} \left (1+r \right ) r \left (-1+r \right )+a_{0} r \left (1+r \right ) \left (-2+r \right )\right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right )+a_{k +1} \left (k +1+r \right ) \left (k +2+r \right ) \left (k +r -1\right )-a_{k} \left (k +1+r \right ) \left (k +r -2\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r \left (-1+r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1, 2\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (\left (a_{k +1}-2 a_{k +2}\right ) k^{2}+\left (\left (2 a_{k +1}-4 a_{k +2}\right ) r -a_{k}+a_{k +1}-4 a_{k +2}\right ) k +\left (a_{k +1}-2 a_{k +2}\right ) r^{2}+\left (-a_{k}+a_{k +1}-4 a_{k +2}\right ) r +2 a_{k}-2 a_{k +1}\right ) \left (k +1+r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}+2 k r a_{k +1}+r^{2} a_{k +1}-k a_{k}+k a_{k +1}-r a_{k}+r a_{k +1}+2 a_{k}-2 a_{k +1}}{2 \left (k^{2}+2 k r +r^{2}+2 k +2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+k a_{k +1}+2 a_{k}-2 a_{k +1}}{2 \left (k^{2}+2 k \right )} \\ \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^{2} a_{k +1}-k a_{k}+k a_{k +1}+2 a_{k}-2 a_{k +1}}{2 \left (k^{2}+2 k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+3 k a_{k +1}+a_{k}}{2 \left (k^{2}+4 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+3 k a_{k +1}+a_{k}}{2 \left (k^{2}+4 k +3\right )}, -2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+5 k a_{k +1}+4 a_{k +1}}{2 \left (k^{2}+6 k +8\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+5 k a_{k +1}+4 a_{k +1}}{2 \left (k^{2}+6 k +8\right )}, -12 a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +2}\right ), a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+3 k a_{k +1}+a_{k}}{2 \left (k^{2}+4 k +3\right )}, -2 a_{0}=0, b_{k +2}=\frac {k^{2} b_{k +1}-k b_{k}+5 k b_{k +1}+4 b_{k +1}}{2 \left (k^{2}+6 k +8\right )}, -12 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 
Louvillian solutions for 3rd order ODEs, imprimitive case: input is reducible, switching to DFactorsols 
checking if the LODE is of Euler type 
expon. solutions partially successful. Result(s) =`, [exp(x), x^2]

\[ y \left (x \right ) = c_{3} \operatorname {expIntegral}_{1}\left (x -2\right ) {\mathrm e}^{x -2}+\frac {c_{3} x^{2} \ln \left (x -2\right )}{4}+{\mathrm e}^{x} c_{2} -\frac {c_{3} x^{2} \ln \left (x \right )}{4}+\frac {\left (2 x +2\right ) c_{3}}{4}+c_{1} x^{2} \]

\[ y(x)\to \frac {1}{4} c_3 \left (-4 e^{x-2} \operatorname {ExpIntegralEi}(2-x)+x^2 \log (2-x)-x^2 \log (x)+2 x+2\right )+c_1 x^2+c_2 e^x \]

4.37 problem 1485

4.37.1 Maple step by step solution