Internal problem ID [9873]
Internal file name [OUTPUT/8820_Monday_June_06_2022_05_33_58_AM_39440404/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1551.
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_high_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {x^{2} y^{\prime \prime \prime \prime }-2 \left (\nu ^{2} x^{2}+6\right ) y^{\prime \prime }+\nu ^{2} \left (\nu ^{2} x^{2}+4\right ) y=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )-2 \left (\nu ^{2} x^{2}+6\right ) \left (\frac {d}{d x}y^{\prime }\right )+\nu ^{2} \left (\nu ^{2} x^{2}+4\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d}{d x}\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 {2 \left (\nu ^{2} x^{2}+6\right )}{x^{2}}, P_{4}\left (x \right )=0, P_{5}\left (x \right )=\frac {\nu ^{2} \left (\nu ^{2} x^{2}+4\right )}{x^{2}}\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}}}=-12 \\ {} & \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^{4}\cdot P_{5}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{4}\cdot P_{5}\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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \nu ^{2} \left (\nu ^{2} x^{2}+4\right ) y+\left (-2 \nu ^{2} x^{2}-12\right ) \left (\frac {d}{d x}y^{\prime }\right )+x^{2} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )=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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} 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 =0..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^{2}\cdot \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}\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 ) \left (k +r -3\right ) x^{k +r -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-2}{\sum }}a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+r \right ) \left (r -6\right ) \left (1+r \right ) x^{-2+r}+a_{1} \left (1+r \right ) r \left (r -5\right ) \left (2+r \right ) x^{-1+r}+\left (a_{2} \left (2+r \right ) \left (1+r \right ) \left (r -4\right ) \left (3+r \right )-2 \nu ^{2} a_{0} \left (1+r \right ) \left (-2+r \right )\right ) x^{r}+\left (a_{3} \left (3+r \right ) \left (2+r \right ) \left (r -3\right ) \left (4+r \right )-2 \nu ^{2} a_{1} \left (2+r \right ) \left (-1+r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (r -4+k \right ) \left (k +3+r \right )-2 \nu ^{2} a_{k} \left (k +1+r \right ) \left (k +r -2\right )+a_{k -2} \nu ^{4}\right ) x^{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 (r -6\right ) \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0, 1, 6\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) r \left (r -5\right ) \left (2+r \right )=0, a_{2} \left (2+r \right ) \left (1+r \right ) \left (r -4\right ) \left (3+r \right )-2 \nu ^{2} a_{0} \left (1+r \right ) \left (-2+r \right )=0, a_{3} \left (3+r \right ) \left (2+r \right ) \left (r -3\right ) \left (4+r \right )-2 \nu ^{2} a_{1} \left (2+r \right ) \left (-1+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=\frac {2 a_{0} \nu ^{2} \left (-2+r \right )}{r^{3}+r^{2}-14 r -24}, a_{3}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \left (r -4+k \right ) \left (k +3+r \right )-2 \nu ^{2} a_{k} \left (k +1+r \right ) \left (k +r -2\right )+a_{k -2} \nu ^{4}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +4} \left (k +4+r \right ) \left (k +3+r \right ) \left (k +r -2\right ) \left (k +5+r \right )-2 \nu ^{2} a_{k +2} \left (k +3+r \right ) \left (k +r \right )+a_{k} \nu ^{4}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=\frac {\nu ^{2} \left (2 k^{2} a_{k +2}+4 k r a_{k +2}-\nu ^{2} a_{k}+2 r^{2} a_{k +2}+6 k a_{k +2}+6 r a_{k +2}\right )}{\left (k +4+r \right ) \left (k +3+r \right ) \left (k +r -2\right ) \left (k +5+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +4}=\frac {\nu ^{2} \left (2 k^{2} a_{k +2}-\nu ^{2} a_{k}+2 k a_{k +2}-4 a_{k +2}\right )}{\left (k +3\right ) \left (k +2\right ) \left (k -3\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-1\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =3 \\ {} & {} & a_{k +4}=\frac {\nu ^{2} \left (2 k^{2} a_{k +2}-\nu ^{2} a_{k}+2 k a_{k +2}-4 a_{k +2}\right )}{\left (k +3\right ) \left (k +2\right ) \left (k -3\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +4}=\frac {\nu ^{2} \left (2 k^{2} a_{k +2}-\nu ^{2} a_{k}+6 k a_{k +2}\right )}{\left (k +4\right ) \left (k +3\right ) \left (k -2\right ) \left (k +5\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 =2 \\ {} & {} & a_{k +4}=\frac {\nu ^{2} \left (2 k^{2} a_{k +2}-\nu ^{2} a_{k}+6 k a_{k +2}\right )}{\left (k +4\right ) \left (k +3\right ) \left (k -2\right ) \left (k +5\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +4}=\frac {\nu ^{2} \left (2 k^{2} a_{k +2}-\nu ^{2} a_{k}+10 k a_{k +2}+8 a_{k +2}\right )}{\left (k +5\right ) \left (k +4\right ) \left (k -1\right ) \left (k +6\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =1\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =1 \\ {} & {} & a_{k +4}=\frac {\nu ^{2} \left (2 k^{2} a_{k +2}-\nu ^{2} a_{k}+10 k a_{k +2}+8 a_{k +2}\right )}{\left (k +5\right ) \left (k +4\right ) \left (k -1\right ) \left (k +6\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =6 \\ {} & {} & a_{k +4}=\frac {\nu ^{2} \left (2 k^{2} a_{k +2}-\nu ^{2} a_{k}+30 k a_{k +2}+108 a_{k +2}\right )}{\left (k +10\right ) \left (k +9\right ) \left (k +4\right ) \left (k +11\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =6 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +6}, a_{k +4}=\frac {\nu ^{2} \left (2 k^{2} a_{k +2}-\nu ^{2} a_{k}+30 k a_{k +2}+108 a_{k +2}\right )}{\left (k +10\right ) \left (k +9\right ) \left (k +4\right ) \left (k +11\right )}, a_{1}=0, a_{2}=\frac {a_{0} \nu ^{2}}{18}, a_{3}=0\right ] \end {array} \]
Maple trace
`Methods for high 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: 4; missing the dependent variable Equation is the LCLM of (x*nu+1)/x*y(x)+diff(y(x),x), ((x*nu+1)/x-(1/5*nu^2*x^2+4/5*x*nu+1)/(1/15*nu^2*x^3+2/5*x^2*nu+x))*y(x)+diff( 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 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 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 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: 62
dsolve(x^2*diff(y(x),x$4)-2*(nu^2*x^2+6)*diff(y(x),x$2)+nu^2*(nu^2*x^2+4)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (c_{4} \nu ^{2} x^{3}+6 c_{4} \nu \,x^{2}+15 c_{4} x +c_{2} \right ) {\mathrm e}^{-x \nu }+{\mathrm e}^{x \nu } \left (c_{3} \nu ^{2} x^{3}-6 c_{3} \nu \,x^{2}+15 c_{3} x +c_{1} \right )}{x} \]
✓ Solution by Mathematica
Time used: 0.197 (sec). Leaf size: 84
DSolve[x^2*y''''[x]-2*(nu^2*x^2+6)*y''[x]+nu^2*(nu^2*x^2+4)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {e^{-\nu x} \left (c_3 \left (-\nu ^2 x^3+\nu ^2-6 \nu x^2+6 \nu -15 x+15\right )+e^{2 \nu x} \left (c_4 \left (-\nu ^2 x^3+\nu ^2+6 \nu x^2-6 \nu -15 x+15\right )+c_2\right )+c_1\right )}{x} \]