2.1.3 problem 4

Internal problem ID [18209]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter 1. section 5. Problems at page 19
Problem number : 4
Date solved : Friday, December 20, 2024 at 05:50:34 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} y^{\prime \prime }+\frac {y^{\prime }}{x}+k^{2} y&=0 \end{align*}

Solved as second order Bessel ode

Time used: 0.057 (sec)

Writing the ode as

\begin{align*} x^{2} y^{\prime \prime }+x y^{\prime }+k^{2} x^{2} 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 &= 0\\ \beta &= k\\ n &= 0\\ \gamma &= 1 \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} y = c_{1} \operatorname {BesselJ}\left (0, k x \right )+c_{2} \operatorname {BesselY}\left (0, k x \right ) \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_{1} \operatorname {BesselJ}\left (0, k x \right )+c_{2} \operatorname {BesselY}\left (0, k x \right ) \\ \end{align*}

Solved as second order ode adjoint method

Time used: 0.746 (sec)

In normal form the ode

\begin{align*} y^{\prime \prime }+\frac {y^{\prime }}{x}+k^{2} y = 0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {1}{x}\\ q \left (x \right )&=k^{2}\\ r \left (x \right )&=0 \end{align*}

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {\xi \left (x \right )}{x}\right )' + \left (k^{2} \xi \left (x \right )\right ) &= 0\\ \frac {\xi \left (x \right ) k^{2} x^{2}+\xi ^{\prime \prime }\left (x \right ) x^{2}-\xi ^{\prime }\left (x \right ) x +\xi \left (x \right )}{x^{2}}&= 0 \end{align*}

Which is solved for \(\xi (x)\). Writing the ode as

\begin{align*} \xi ^{\prime \prime } x^{2}-\xi ^{\prime } x +\left (k^{2} x^{2}+1\right ) \xi = 0\tag {1} \end{align*}

Bessel ode has the form

\begin{align*} \xi ^{\prime \prime } x^{2}+\xi ^{\prime } x +\left (-n^{2}+x^{2}\right ) \xi = 0\tag {2} \end{align*}

The generalized form of Bessel ode is given by Bowman (1958) as the following

\begin{align*} \xi ^{\prime \prime } x^{2}+\left (1-2 \alpha \right ) x \xi ^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) \xi = 0\tag {3} \end{align*}

With the standard solution

\begin{align*} \xi &=x^{\alpha } \left (c_3 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_4 \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 &= k\\ n &= 0\\ \gamma &= 1 \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} \xi = c_3 x \operatorname {BesselJ}\left (0, k x \right )+c_4 x \operatorname {BesselY}\left (0, k x \right ) \end{align*}

Will add steps showing solving for IC soon.

The original ode now reduces to first order ode

\begin{align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )} \end{align*}

Or

\begin{align*} y^{\prime }+y \left (\frac {1}{x}-\frac {c_3 \operatorname {BesselJ}\left (0, k x \right )-c_3 x \operatorname {BesselJ}\left (1, k x \right ) k +c_4 \operatorname {BesselY}\left (0, k x \right )-c_4 x \operatorname {BesselY}\left (1, k x \right ) k}{c_3 x \operatorname {BesselJ}\left (0, k x \right )+c_4 x \operatorname {BesselY}\left (0, k x \right )}\right )&=0 \end{align*}

Which is now a first order ode. This is now solved for \(y\). In canonical form a linear first order is

\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\frac {k \left (\operatorname {BesselJ}\left (1, k x \right ) c_3 +\operatorname {BesselY}\left (1, k x \right ) c_4 \right )}{c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right )}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {k \left (\operatorname {BesselJ}\left (1, k x \right ) c_3 +\operatorname {BesselY}\left (1, k x \right ) c_4 \right )}{c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right )}d x}\\ &= \frac {1}{c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right )} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right )}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} \frac {y}{c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right )}&= \int {0 \,dx} + c_5 \\ &=c_5 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right )}\) gives the final solution

\[ y = \left (c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right )\right ) c_5 \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \left (c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right )\right ) c_5 \\ \end{align*}

The constants can be merged to give

\[ y = c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right ) \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_3 \operatorname {BesselJ}\left (0, k x \right )+c_4 \operatorname {BesselY}\left (0, k x \right ) \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime }}{x}+k^{2} y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & 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 )=k^{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}}}=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 {Multiply by denominators}\hspace {3pt} \\ {} & {} & k^{2} y x +y^{\prime \prime } x +y^{\prime }=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 \cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\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 y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r^{2} x^{-1+r}+a_{1} \left (1+r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right )^{2}+k^{2} a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right )^{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1\right )^{2}+k^{2} a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2\right )^{2}+k^{2} a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}}{\left (k +2\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}}{\left (k +2\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {k^{2} a_{k}}{\left (k +2\right )^{2}}, a_{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`
 

Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 19

dsolve(diff(diff(y(x),x),x)+1/x*diff(y(x),x)+k^2*y(x) = 0, 
       y(x),singsol=all)
 

\[ y = c_{1} \operatorname {BesselJ}\left (0, k x \right )+c_{2} \operatorname {BesselY}\left (0, k x \right ) \]

Mathematica DSolve solution

Solving time : 0.029 (sec)
Leaf size : 22

DSolve[{D[y[x],{x,2}]+1/x*D[y[x],x]+k^2*y[x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to c_1 \operatorname {BesselJ}(0,k x)+c_2 \operatorname {BesselY}(0,k x) \]