Internal problem ID [9816]
Internal file name [OUTPUT/8759_Monday_June_06_2022_05_24_32_AM_20221889/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1490.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_missing_y"
Maple gives the following as the ode type
[[_3rd_order, _missing_y]]
\[ \boxed {x^{2} y^{\prime \prime \prime }-y^{\prime \prime } x +\left (x^{2}+1\right ) y^{\prime }=0} \] Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} x^{2} v^{\prime \prime }\left (x \right )-v^{\prime }\left (x \right ) x +\left (x^{2}+1\right ) v \left (x \right ) = 0 \end {align*}
Writing the ode as \begin {align*} x^{2} v^{\prime \prime }\left (x \right )-v^{\prime }\left (x \right ) x +\left (x^{2}+1\right ) v \left (x \right ) = 0\tag {1} \end {align*}
Bessel ode has the form \begin {align*} x^{2} v^{\prime \prime }\left (x \right )+v^{\prime }\left (x \right ) x +\left (-n^{2}+x^{2}\right ) v \left (x \right ) = 0\tag {2} \end {align*}
The generalized form of Bessel ode is given by Bowman (1958) as the following \begin {align*} x^{2} v^{\prime \prime }\left (x \right )+\left (1-2 \alpha \right ) x v^{\prime }\left (x \right )+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) v \left (x \right ) = 0\tag {3} \end {align*}
With the standard solution \begin {align*} v \left (x \right )&=x^{\alpha } \left (c_{1} \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_{2} \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end {align*}
Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives \begin {align*} \alpha &= 1\\ \beta &= 1\\ n &= 0\\ \gamma &= 1 \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} v \left (x \right ) = c_{1} x \operatorname {BesselJ}\left (0, x\right )+c_{2} x \operatorname {BesselY}\left (0, x\right ) \end {align*}
But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = c_{1} x \operatorname {BesselJ}\left (0, x\right )+c_{2} x \operatorname {BesselY}\left (0, x\right )\). Integrating both sides gives \begin {align*} y &= \int { c_{1} x \operatorname {BesselJ}\left (0, x\right )+c_{2} x \operatorname {BesselY}\left (0, x\right )\,\mathop {\mathrm {d}x}}\\ &= c_{1} x \operatorname {BesselJ}\left (1, x\right )+c_{2} x \operatorname {BesselY}\left (1, x\right )+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \operatorname {BesselJ}\left (1, x\right )+c_{2} x \operatorname {BesselY}\left (1, x\right )+c_{3} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \operatorname {BesselJ}\left (1, x\right )+c_{2} x \operatorname {BesselY}\left (1, x\right )+c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime \prime }\right )-\left (\frac {d}{d x}y^{\prime }\right ) x +\left (x^{2}+1\right ) y^{\prime }=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 {1}{x}, P_{3}\left (x \right )=\frac {x^{2}+1}{x^{2}}, P_{4}\left (x \right )=0\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}}}=-1 \\ {} & \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}}}=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}}}=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^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & 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 )^{2} x^{-1+r}+a_{1} \left (1+r \right ) \left (-1+r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )^{2}+a_{k -1} \left (k +r -1\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 )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (-1+r \right )^{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -1\right ) \left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )+a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +r \right ) \left (a_{k +2} \left (k +2+r \right ) \left (k +r \right )+a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +r \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{k \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 +2}=-\frac {a_{k}}{k \left (k +2\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2\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 +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +4\right )}, 3 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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +2}\right ), a_{k +2}=-\frac {a_{k}}{k \left (k +2\right )}, a_{1}=0, b_{k +2}=-\frac {b_{k}}{\left (k +2\right ) \left (k +4\right )}, 3 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 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 <- missing the dependent variable successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 18
dsolve(x^2*diff(diff(diff(y(x),x),x),x)-x*diff(diff(y(x),x),x)+(x^2+1)*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} +c_{2} x \operatorname {BesselJ}\left (1, x\right )+c_{3} x \operatorname {BesselY}\left (1, x\right ) \]
✓ Solution by Mathematica
Time used: 0.041 (sec). Leaf size: 33
DSolve[(1 + x^2)*y'[x] - x*y''[x] + x^2*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} c_1 x^2 \operatorname {Hypergeometric0F1Regularized}\left (2,-\frac {x^2}{4}\right )+c_2 x \operatorname {BesselY}(1,x)+c_3 \]