4.73 problem 1523

Internal problem ID [9847]
Internal file name [OUTPUT/8792_Monday_June_06_2022_05_29_48_AM_37314478/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1523.
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.

\[ \boxed {\left (x^{2}+1\right ) x^{3} y^{\prime \prime \prime }-\left (4 x^{2}+2\right ) x^{2} y^{\prime \prime }+\left (10 x^{2}+4\right ) x y^{\prime }-4 \left (3 x^{2}+1\right ) y=0} \] Unable to solve this ODE.

4.73.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-\left (4 x^{2}+2\right ) x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (10 x^{2}+4\right ) x y^{\prime }-4 \left (3 x^{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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 \left (2 x^{2}+1\right )}{\left (x^{2}+1\right ) x}, P_{3}\left (x \right )=\frac {2 \left (5 x^{2}+2\right )}{\left (x^{2}+1\right ) x^{2}}, P_{4}\left (x \right )=-\frac {4 \left (3 x^{2}+1\right )}{\left (x^{2}+1\right ) 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}}}=-2 \\ {} & \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 \\ {} & \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 \\ {} & \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 {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )-2 \left (2 x^{2}+1\right ) x^{2} \left (\frac {d}{d x}y^{\prime }\right )+2 x \left (5 x^{2}+2\right ) y^{\prime }+\left (-12 x^{2}-4\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\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 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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & 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 =3..5 \\ {} & {} & 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} \\ {} & {} & a_{0} \left (r -1\right ) \left (r -2\right )^{2} x^{r}+a_{1} r \left (r -1\right )^{2} x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r -1\right ) \left (k +r -2\right )^{2}+a_{k -2} \left (k -5+r \right ) \left (k -4+r \right )^{2}\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 -2\right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, 2\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} r \left (r -1\right )^{2}=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 (k +r -2\right )^{2}+a_{k -2} \left (k -5+r \right ) \left (k -4+r \right )^{2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +1+r \right ) \left (k +r \right )^{2}+a_{k} \left (k +r -3\right ) \left (k +r -2\right )^{2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k +r -3\right ) \left (k +r -2\right )^{2}}{\left (k +1+r \right ) \left (k +r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k -2\right ) \left (k -1\right )^{2}}{\left (k +2\right ) \left (k +1\right )^{2}} \\ \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 {a_{k} \left (k -2\right ) \left (k -1\right )^{2}}{\left (k +2\right ) \left (k +1\right )^{2}}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k -1\right ) k^{2}}{\left (k +3\right ) \left (k +2\right )^{2}} \\ \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 {a_{k} \left (k -1\right ) k^{2}}{\left (k +3\right ) \left (k +2\right )^{2}}, 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 {a_{k} \left (k -2\right ) \left (k -1\right )^{2}}{\left (k +2\right ) \left (k +1\right )^{2}}, a_{1}=0, b_{k +2}=-\frac {b_{k} \left (k -1\right ) k^{2}}{\left (k +3\right ) \left (k +2\right )^{2}}, 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) =`, [x^2, x^3+x]
 

✓ Solution by Maple

Time used: 0.062 (sec). Leaf size: 23

dsolve((x^2+1)*x^3*diff(diff(diff(y(x),x),x),x)-(4*x^2+2)*x^2*diff(diff(y(x),x),x)+(10*x^2+4)*x*diff(y(x),x)-4*(3*x^2+1)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = x \left (c_{2} x \ln \left (x \right )+x^{2} c_{3} +\left (c_{1} +c_{2} \right ) x +c_{3} \right ) \]

✓ Solution by Mathematica

Time used: 0.495 (sec). Leaf size: 46

DSolve[-4*(1 + 3*x^2)*y[x] + x*(4 + 10*x^2)*y'[x] - x^2*(2 + 4*x^2)*y''[x] + x^3*(1 + x^2)*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{2} x \left (c_2 x^2-2 c_1 \left (x^2-3 x+1\right )-2 c_2 x+c_3 x+c_3 x \log (x)+c_2\right ) \]