Internal problem ID [15489]
Internal file name [OUTPUT/15490_Friday_May_10_2024_05_47_27_PM_88000253/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 18.2. Expanding a solution in generalized power
series. Bessels equation. Exercises page 177
Problem number: 748.
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]]
\[ \boxed {x^{2} y^{\prime \prime }-2 x y^{\prime }+4 \left (x^{4}-1\right ) y=0} \]
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }-2 x y^{\prime }+\left (4 x^{4}-4\right ) y = 0\tag {1} \end {align*}
Bessel ode has the form \begin {align*} x^{2} y^{\prime \prime }+x y^{\prime }+\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 {3}{2}}\\ \beta &= 1\\ n &= -{\frac {5}{4}}\\ \gamma &= 2 \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} y = c_{1} x^{\frac {3}{2}} \operatorname {BesselJ}\left (-\frac {5}{4}, x^{2}\right )+c_{2} x^{\frac {3}{2}} \operatorname {BesselY}\left (-\frac {5}{4}, x^{2}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {3}{2}} \operatorname {BesselJ}\left (-\frac {5}{4}, x^{2}\right )+c_{2} x^{\frac {3}{2}} \operatorname {BesselY}\left (-\frac {5}{4}, x^{2}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {3}{2}} \operatorname {BesselJ}\left (-\frac {5}{4}, x^{2}\right )+c_{2} x^{\frac {3}{2}} \operatorname {BesselY}\left (-\frac {5}{4}, x^{2}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }-2 x y^{\prime }+\left (4 x^{4}-4\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {4 \left (x^{4}-1\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} \\ {} & {} & y^{\prime \prime }-\frac {2 y^{\prime }}{x}+\frac {4 \left (x^{4}-1\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 {4 \left (x^{4}-1\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}}}=-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 =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} y^{\prime \prime }-2 x y^{\prime }+\left (4 x^{4}-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..4 \\ {} & {} & 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 y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\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 (1+r \right ) \left (-4+r \right ) x^{r}+a_{1} \left (2+r \right ) \left (-3+r \right ) x^{1+r}+a_{2} \left (3+r \right ) \left (-2+r \right ) x^{2+r}+a_{3} \left (4+r \right ) \left (-1+r \right ) x^{3+r}+\left (\moverset {\infty }{\munderset {k =4}{\sum }}\left (a_{k} \left (k +r +1\right ) \left (k +r -4\right )+4 a_{k -4}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+r \right ) \left (-4+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 4\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (2+r \right ) \left (-3+r \right )=0, a_{2} \left (3+r \right ) \left (-2+r \right )=0, a_{3} \left (4+r \right ) \left (-1+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} \\ {} & {} & a_{k} \left (k +r +1\right ) \left (k +r -4\right )+4 a_{k -4}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +4 \\ {} & {} & a_{k +4} \left (k +5+r \right ) \left (k +r \right )+4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +5+r \right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +4\right ) \left (k -1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +4}=-\frac {4 a_{k}}{\left (k +4\right ) \left (k -1\right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =4 \\ {} & {} & a_{k +4}=-\frac {4 a_{k}}{\left (k +9\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =4 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +4}, a_{k +4}=-\frac {4 a_{k}}{\left (k +9\right ) \left (k +4\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 -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +4}\right ), a_{k +4}=-\frac {4 a_{k}}{\left (k +4\right ) \left (k -1\right )}, a_{1}=0, a_{2}=0, a_{3}=0, b_{k +4}=-\frac {4 b_{k}}{\left (k +9\right ) \left (k +4\right )}, b_{1}=0, b_{2}=0, b_{3}=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: 0.0 (sec). Leaf size: 45
dsolve(x^2*diff(y(x),x$2)-2*x*diff(y(x),x)+4*(x^4-1)*y(x)=0,y(x), singsol=all)
\[ y = -\frac {-\frac {\operatorname {BesselY}\left (\frac {1}{4}, x^{2}\right ) c_{2}}{2}-\frac {\operatorname {BesselJ}\left (\frac {1}{4}, x^{2}\right ) c_{1}}{2}+x^{2} \left (c_{1} \operatorname {BesselJ}\left (-\frac {3}{4}, x^{2}\right )+\operatorname {BesselY}\left (-\frac {3}{4}, x^{2}\right ) c_{2} \right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.158 (sec). Leaf size: 46
DSolve[x^2*y''[x]-2*x*y'[x]+4*(x^4-1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x^{3/2} \left (c_2 \operatorname {Gamma}\left (\frac {9}{4}\right ) \operatorname {BesselJ}\left (\frac {5}{4},x^2\right )-4 c_1 \operatorname {Gamma}\left (\frac {3}{4}\right ) \operatorname {BesselJ}\left (-\frac {5}{4},x^2\right )\right )}{2^{3/4}} \]