3.4 problem 4

3.4.1 Solving as second order bessel ode ode
3.4.2 Maple step by step solution

Internal problem ID [6622]
Internal file name [OUTPUT/5870_Sunday_June_05_2022_03_59_07_PM_68484540/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS. 6.4 SPECIAL FUNCTIONS. EXERCISES 6.4. Page 267
Problem number: 4.
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]]

\[ \boxed {16 x^{2} y^{\prime \prime }+16 y^{\prime } x +\left (16 x^{2}-1\right ) y=0} \]

3.4.1 Solving as second order bessel ode ode

Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (x^{2}-\frac {1}{16}\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 &= 0\\ \beta &= 1\\ n &= -{\frac {1}{4}}\\ \gamma &= 1 \end {align*}

Substituting all the above into (4) gives the solution as \begin {align*} y = c_{1} \operatorname {BesselJ}\left (-\frac {1}{4}, x\right )+c_{2} \operatorname {BesselY}\left (-\frac {1}{4}, x\right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \operatorname {BesselJ}\left (-\frac {1}{4}, x\right )+c_{2} \operatorname {BesselY}\left (-\frac {1}{4}, x\right ) \\ \end{align*}

Verification of solutions

\[ y = c_{1} \operatorname {BesselJ}\left (-\frac {1}{4}, x\right )+c_{2} \operatorname {BesselY}\left (-\frac {1}{4}, x\right ) \] Verified OK.

3.4.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 16 x^{2} y^{\prime \prime }+16 y^{\prime } x +\left (16 x^{2}-1\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 {\left (16 x^{2}-1\right ) y}{16 x^{2}}-\frac {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 {y^{\prime }}{x}+\frac {\left (16 x^{2}-1\right ) y}{16 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 {1}{x}, P_{3}\left (x \right )=\frac {16 x^{2}-1}{16 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}}}=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}}}=-\frac {1}{16} \\ {} & \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} \\ {} & {} & 16 x^{2} y^{\prime \prime }+16 y^{\prime } x +\left (16 x^{2}-1\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 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+4 r \right ) \left (-1+4 r \right ) x^{r}+a_{1} \left (5+4 r \right ) \left (3+4 r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (4 k +4 r +1\right ) \left (4 k +4 r -1\right )+16 a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+4 r \right ) \left (-1+4 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{4}, \frac {1}{4}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (5+4 r \right ) \left (3+4 r \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} \\ {} & {} & a_{k} \left (4 k +4 r +1\right ) \left (4 k +4 r -1\right )+16 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (4 k +9+4 r \right ) \left (4 k +7+4 r \right )+16 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {16 a_{k}}{\left (4 k +9+4 r \right ) \left (4 k +7+4 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{4} \\ {} & {} & a_{k +2}=-\frac {16 a_{k}}{\left (4 k +8\right ) \left (4 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{4}}, a_{k +2}=-\frac {16 a_{k}}{\left (4 k +8\right ) \left (4 k +6\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{4} \\ {} & {} & a_{k +2}=-\frac {16 a_{k}}{\left (4 k +10\right ) \left (4 k +8\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{4}}, a_{k +2}=-\frac {16 a_{k}}{\left (4 k +10\right ) \left (4 k +8\right )}, 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 -\frac {1}{4}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{4}}\right ), a_{k +2}=-\frac {16 a_{k}}{\left (4 k +8\right ) \left (4 k +6\right )}, a_{1}=0, b_{k +2}=-\frac {16 b_{k}}{\left (4 k +10\right ) \left (4 k +8\right )}, b_{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: 0.016 (sec). Leaf size: 15

dsolve(16*x^2*diff(y(x),x$2)+16*x*diff(y(x),x)+(16*x^2-1)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = c_{1} \operatorname {BesselJ}\left (\frac {1}{4}, x\right )+c_{2} \operatorname {BesselY}\left (\frac {1}{4}, x\right ) \]

Solution by Mathematica

Time used: 0.017 (sec). Leaf size: 22

DSolve[16*x^2*y''[x]+16*x*y'[x]+(16*x^2-1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_1 \operatorname {BesselJ}\left (\frac {1}{4},x\right )+c_2 \operatorname {BesselY}\left (\frac {1}{4},x\right ) \]