Internal problem ID [9499]
Internal file name [OUTPUT/8439_Monday_June_06_2022_03_04_39_AM_4148697/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1170.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_bessel_ode"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+2 y^{\prime } x +\left (a \,x^{2}+b \right ) y=0} \]
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+2 y^{\prime } x +\left (a \,x^{2}+b \right ) y = 0\tag {1} \end {align*}
Bessel ode has the form \begin {align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (-n^{2}+x^{2}\right ) y = 0\tag {2} \end {align*}
The generalized form of Bessel ode is given by Bowman (1958) as the following \begin {align*} x^{2} y^{\prime \prime }+\left (1-2 \alpha \right ) x y^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) y = 0\tag {3} \end {align*}
With the standard solution \begin {align*} y&=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 &= -{\frac {1}{2}}\\ \beta &= \sqrt {a}\\ n &= \frac {\sqrt {-4 b +1}}{2}\\ \gamma &= 1 \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} y = \frac {c_{1} \operatorname {BesselJ}\left (\frac {\sqrt {-4 b +1}}{2}, \sqrt {a}\, x \right )}{\sqrt {x}}+\frac {c_{2} \operatorname {BesselY}\left (\frac {\sqrt {-4 b +1}}{2}, \sqrt {a}\, x \right )}{\sqrt {x}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \operatorname {BesselJ}\left (\frac {\sqrt {-4 b +1}}{2}, \sqrt {a}\, x \right )}{\sqrt {x}}+\frac {c_{2} \operatorname {BesselY}\left (\frac {\sqrt {-4 b +1}}{2}, \sqrt {a}\, x \right )}{\sqrt {x}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \operatorname {BesselJ}\left (\frac {\sqrt {-4 b +1}}{2}, \sqrt {a}\, x \right )}{\sqrt {x}}+\frac {c_{2} \operatorname {BesselY}\left (\frac {\sqrt {-4 b +1}}{2}, \sqrt {a}\, x \right )}{\sqrt {x}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+2 y^{\prime } x +\left (a \,x^{2}+b \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (a \,x^{2}+b \right ) y}{x^{2}}-\frac {2 y^{\prime }}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {2 y^{\prime }}{x}+\frac {\left (a \,x^{2}+b \right ) y}{x^{2}}=0 \\ \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 {2}{x}, P_{3}\left (x \right )=\frac {a \,x^{2}+b}{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}}}=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}}}=b \\ {} & \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^{2} \left (\frac {d}{d x}y^{\prime }\right )+2 y^{\prime } x +\left (a \,x^{2}+b \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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r^{2}+b +r \right ) x^{r}+a_{1} \left (r^{2}+b +3 r +2\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}+b +k +r \right )+a_{k -2} a \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}+b +r =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2}, -\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (r^{2}+b +3 r +2\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} \\ {} & {} & \left (k^{2}+\left (2 r +1\right ) k +r^{2}+b +r \right ) a_{k}+a_{k -2} a =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+\left (2 r +1\right ) \left (k +2\right )+r^{2}+b +r \right ) a_{k +2}+a_{k} a =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} a}{k^{2}+2 k r +r^{2}+b +5 k +5 r +6} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2} \\ {} & {} & a_{k +2}=-\frac {a_{k} a}{k^{2}+2 k \left (-\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2}\right )+\left (-\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2}\right )^{2}+b +5 k +\frac {7}{2}-\frac {5 \sqrt {-4 b +1}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2}}, a_{k +2}=-\frac {a_{k} a}{k^{2}+2 k \left (-\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2}\right )+\left (-\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2}\right )^{2}+b +5 k +\frac {7}{2}-\frac {5 \sqrt {-4 b +1}}{2}}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2} \\ {} & {} & a_{k +2}=-\frac {a_{k} a}{k^{2}+2 k \left (-\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2}\right )+\left (-\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2}\right )^{2}+b +5 k +\frac {7}{2}+\frac {5 \sqrt {-4 b +1}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2}}, a_{k +2}=-\frac {a_{k} a}{k^{2}+2 k \left (-\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2}\right )+\left (-\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2}\right )^{2}+b +5 k +\frac {7}{2}+\frac {5 \sqrt {-4 b +1}}{2}}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k -\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k -\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2}}\right ), c_{k +2}=-\frac {c_{k} a}{k^{2}+2 k \left (-\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2}\right )+\left (-\frac {1}{2}-\frac {\sqrt {-4 b +1}}{2}\right )^{2}+b +5 k +\frac {7}{2}-\frac {5 \sqrt {-4 b +1}}{2}}, c_{1}=0, d_{k +2}=-\frac {d_{k} a}{k^{2}+2 k \left (-\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2}\right )+\left (-\frac {1}{2}+\frac {\sqrt {-4 b +1}}{2}\right )^{2}+b +5 k +\frac {7}{2}+\frac {5 \sqrt {-4 b +1}}{2}}, d_{1}=0\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- 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: 0.047 (sec). Leaf size: 43
dsolve(x^2*diff(diff(y(x),x),x)+2*x*diff(y(x),x)+(a*x^2+b)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{1} \operatorname {BesselJ}\left (\frac {\sqrt {1-4 b}}{2}, x \sqrt {a}\right )+c_{2} \operatorname {BesselY}\left (\frac {\sqrt {1-4 b}}{2}, x \sqrt {a}\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.04 (sec). Leaf size: 58
DSolve[(b + a*x^2)*y[x] + 2*x*y'[x] + x^2*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 j_{\frac {1}{2} \left (\sqrt {1-4 b}-1\right )}\left (\sqrt {a} x\right )+c_2 y_{\frac {1}{2} \left (\sqrt {1-4 b}-1\right )}\left (\sqrt {a} x\right ) \]