Internal problem ID [11329]
Internal file name [OUTPUT/10315_Tuesday_December_27_2022_04_07_29_AM_43734921/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter IX, Miscellaneous methods for solving equations of higher order than first.
Article 60. Exact equation. Integrating factor. Page 139
Problem number: Ex 8.
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.
\[ \boxed {x^{2} y^{\prime \prime \prime }-5 x y^{\prime \prime }+\left (4 x^{4}+5\right ) y^{\prime }-8 y x^{3}=0} \] Unable to solve this ODE.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime \prime }\right ) x^{2}-5 \left (\frac {d}{d x}y^{\prime }\right ) x +\left (4 x^{4}+5\right ) y^{\prime }-8 y x^{3}=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 )=-\frac {5}{x}, P_{3}\left (x \right )=\frac {4 x^{4}+5}{x^{2}}, P_{4}\left (x \right )=-8 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}}}=-5 \\ {} & \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}}}=5 \\ {} & \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}=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} x^{3}\cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +3} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -3 \\ {} & {} & x^{3}\cdot y=\moverset {\infty }{\munderset {k =3}{\sum }}a_{k -3} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..4 \\ {} & {} & 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 \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime \prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \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 (-2+r \right ) \left (-6+r \right ) x^{-1+r}+a_{1} \left (1+r \right ) \left (-1+r \right ) \left (-5+r \right ) x^{r}+a_{2} \left (2+r \right ) r \left (-4+r \right ) x^{1+r}+a_{3} \left (3+r \right ) \left (1+r \right ) \left (-3+r \right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right ) \left (k -5+r \right )+4 a_{k -3} \left (k -5+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-2+r \right ) \left (-6+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2, 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 ) \left (-1+r \right ) \left (-5+r \right )=0, a_{2} \left (2+r \right ) r \left (-4+r \right )=0, a_{3} \left (3+r \right ) \left (1+r \right ) \left (-3+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=0, a_{3}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k -5+r \right ) \left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )+4 a_{k -3}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & \left (k +r -2\right ) \left (a_{k +4} \left (k +4+r \right ) \left (k +2+r \right )+4 a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +4+r \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +4\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=-\frac {4 a_{k}}{\left (k +4\right ) \left (k +2\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +6\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +4}=-\frac {4 a_{k}}{\left (k +6\right ) \left (k +4\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =6 \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +10\right ) \left (k +8\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 {4 a_{k}}{\left (k +10\right ) \left (k +8\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +6}\right ), a_{k +4}=-\frac {4 a_{k}}{\left (k +4\right ) \left (k +2\right )}, a_{1}=0, a_{2}=0, a_{3}=0, b_{k +4}=-\frac {4 b_{k}}{\left (k +6\right ) \left (k +4\right )}, b_{1}=0, b_{2}=0, b_{3}=0, c_{k +4}=-\frac {4 c_{k}}{\left (k +10\right ) \left (k +8\right )}, c_{1}=0, c_{2}=0, c_{3}=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 -2/x*y(x)+diff(y(x),x), 4*x^2*y(x)-1/x*diff(y(x),x)+diff(diff(y(x),x),x) 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 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)] <- linear_1 successful <- solving the LCLM ode successful `
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 22
dsolve(x^2*diff(y(x),x$3)-5*x*diff(y(x),x$2)+(4*x^4+5)*diff(y(x),x)-8*x^3*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} x^{2}+c_{2} \cos \left (x^{2}\right )+c_{3} \sin \left (x^{2}\right ) \]
✓ Solution by Mathematica
Time used: 0.507 (sec). Leaf size: 44
DSolve[x^2*y'''[x]-5*x*y''[x]+(4*x^4+5)*y'[x]-8*x^3*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 x^2+\frac {1}{2} i c_2 e^{-i x^2}-\frac {1}{8} c_3 e^{i x^2} \]