Internal problem ID [9833]
Internal file name [OUTPUT/8778_Monday_June_06_2022_05_26_51_AM_11473947/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1509.
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.
\[ \boxed {x^{3} y^{\prime \prime \prime }+\left (4 x^{3}+\left (-4 \nu ^{2}+1\right ) x \right ) y^{\prime }+\left (4 \nu ^{2}-1\right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )+\left (4 x^{3}+\left (-4 \nu ^{2}+1\right ) x \right ) y^{\prime }+\left (4 \nu ^{2}-1\right ) 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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=0, P_{3}\left (x \right )=-\frac {4 \nu ^{2}-4 x^{2}-1}{x^{2}}, P_{4}\left (x \right )=\frac {4 \nu ^{2}-1}{x^{3}}\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}}}=-4 \nu ^{2}+1 \\ {} & \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}}}=4 \nu ^{2}-1 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-x \left (4 \nu ^{2}-4 x^{2}-1\right ) y^{\prime }+\left (4 \nu ^{2}-1\right ) y=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} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{3}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r -1\right ) \left (r -1+2 \nu \right ) \left (r -1-2 \nu \right ) x^{r}+a_{1} r \left (r +2 \nu \right ) \left (r -2 \nu \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r -1\right ) \left (r -1+2 \nu +k \right ) \left (r -1-2 \nu +k \right )+4 a_{k -2} \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} \\ {} & {} & \left (r -1\right ) \left (r -1+2 \nu \right ) \left (r -1-2 \nu \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, 1-2 \nu , 2 \nu +1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} r \left (r +2 \nu \right ) \left (r -2 \nu \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r -1\right ) \left (r -1+2 \nu +k \right ) \left (r -1-2 \nu +k \right )+4 a_{k -2} \left (k +r -2\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +1+r \right ) \left (r +1+2 \nu +k \right ) \left (r +1-2 \nu +k \right )+4 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {4 a_{k} \left (k +r \right )}{\left (k +1+r \right ) \left (r +1+2 \nu +k \right ) \left (r +1-2 \nu +k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {4 a_{k} \left (k +1\right )}{\left (k +2\right ) \left (2+2 \nu +k \right ) \left (2-2 \nu +k \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 {4 a_{k} \left (k +1\right )}{\left (k +2\right ) \left (2+2 \nu +k \right ) \left (2-2 \nu +k \right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1-2 \nu \\ {} & {} & a_{k +2}=-\frac {4 a_{k} \left (k +1-2 \nu \right )}{\left (2-2 \nu +k \right ) \left (k +2\right ) \left (2-4 \nu +k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1-2 \nu \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1-2 \nu }, a_{k +2}=-\frac {4 a_{k} \left (k +1-2 \nu \right )}{\left (2-2 \nu +k \right ) \left (k +2\right ) \left (2-4 \nu +k \right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \nu +1 \\ {} & {} & a_{k +2}=-\frac {4 a_{k} \left (k +2 \nu +1\right )}{\left (2+2 \nu +k \right ) \left (4 \nu +2+k \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \nu +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2 \nu +1}, a_{k +2}=-\frac {4 a_{k} \left (k +2 \nu +1\right )}{\left (2+2 \nu +k \right ) \left (4 \nu +2+k \right ) \left (k +2\right )}, 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 +1-2 \nu }\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +2 \nu +1}\right ), a_{k +2}=-\frac {4 a_{k} \left (k +1\right )}{\left (k +2\right ) \left (2+2 \nu +k \right ) \left (2-2 \nu +k \right )}, a_{1}=0, b_{k +2}=-\frac {4 b_{k} \left (k +1-2 \nu \right )}{\left (2-2 \nu +k \right ) \left (k +2\right ) \left (2-4 \nu +k \right )}, b_{1}=0, c_{k +2}=-\frac {4 c_{k} \left (k +2 \nu +1\right )}{\left (2+2 \nu +k \right ) \left (4 \nu +2+k \right ) \left (k +2\right )}, c_{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 Equation is the 2nd symmetric power of diff(diff(y(x),x),x)-1/4/x^2*(4*nu^2-4*x^2-1)*y(x) = 0 -> Attempting now to solve this lower order ODE 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 <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 29
dsolve(x^3*diff(diff(diff(y(x),x),x),x)+(4*x^3+(-4*nu^2+1)*x)*diff(y(x),x)+(4*nu^2-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x \left (c_{1} \operatorname {BesselJ}\left (\nu , x\right )^{2}+c_{2} \operatorname {BesselY}\left (\nu , x\right )^{2}+c_{3} \operatorname {BesselJ}\left (\nu , x\right ) \operatorname {BesselY}\left (\nu , x\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 33
DSolve[(-1 + 4*nu^2)*y[x] + ((1 - 4*nu^2)*x + 4*x^3)*y'[x] + x^3*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x \left (c_1 \operatorname {BesselJ}(\nu ,x)^2+c_3 \operatorname {BesselY}(\nu ,x)^2+c_2 \operatorname {BesselJ}(\nu ,x) \operatorname {BesselY}(\nu ,x)\right ) \]