2.45 problem 45

2.45.1 Solved as second order Bessel ode
2.45.2 Solved as second order ode adjoint method
2.45.3 Maple step by step solution
2.45.4 Maple trace
2.45.5 Maple dsolve solution
2.45.6 Mathematica DSolve solution

Internal problem ID [8134]
Book : Second order enumerated odes
Section : section 2
Problem number : 45
Date solved : Monday, October 21, 2024 at 04:54:14 PM
CAS classification : [_Bessel]

Solve

\begin{align*} x^{2} y^{\prime \prime }+x y^{\prime }+\left (x^{2}-5\right ) y&=0 \end{align*}

2.45.1 Solved as second order Bessel ode

Time used: 0.052 (sec)

Writing the ode as

\begin{align*} x^{2} y^{\prime \prime }+x y^{\prime }+\left (x^{2}-5\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 &= 0\\ \beta &= 1\\ n &= -\sqrt {5}\\ \gamma &= 1 \end{align*}

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

\begin{align*} y = c_1 \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_2 \operatorname {BesselY}\left (-\sqrt {5}, x\right ) \end{align*}

Will add steps showing solving for IC soon.

2.45.2 Solved as second order ode adjoint method

Time used: 0.771 (sec)

In normal form the ode

\begin{align*} x^{2} y^{\prime \prime }+x y^{\prime }+\left (x^{2}-5\right ) 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 )&=\frac {x^{2}-5}{x^{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 (\frac {\left (x^{2}-5\right ) \xi \left (x \right )}{x^{2}}\right ) &= 0\\ \frac {\xi ^{\prime \prime }\left (x \right ) x^{2}+\xi \left (x \right ) x^{2}-\xi ^{\prime }\left (x \right ) x -4 \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 (x^{2}-4\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 &= 1\\ n &= -\sqrt {5}\\ \gamma &= 1 \end{align*}

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

\begin{align*} \xi = c_3 x \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_4 x \operatorname {BesselY}\left (-\sqrt {5}, x\right ) \end{align*}

Will add steps showing solving for IC soon.

The original ode (2) 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 )}\\ y^{\prime }+y \left (\frac {1}{x}-\frac {c_3 \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_3 x \left (-\operatorname {BesselJ}\left (-\sqrt {5}+1, x\right )-\frac {\sqrt {5}\, \operatorname {BesselJ}\left (-\sqrt {5}, x\right )}{x}\right )+c_4 \operatorname {BesselY}\left (-\sqrt {5}, x\right )+c_4 x \left (-\operatorname {BesselY}\left (-\sqrt {5}+1, x\right )-\frac {\sqrt {5}\, \operatorname {BesselY}\left (-\sqrt {5}, x\right )}{x}\right )}{c_3 x \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_4 x \operatorname {BesselY}\left (-\sqrt {5}, 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 {\operatorname {BesselJ}\left (-\sqrt {5}, x\right ) \sqrt {5}\, c_3 +\operatorname {BesselJ}\left (-\sqrt {5}+1, x\right ) c_3 x +\sqrt {5}\, \operatorname {BesselY}\left (-\sqrt {5}, x\right ) c_4 +\operatorname {BesselY}\left (-\sqrt {5}+1, x\right ) c_4 x}{x \left (c_3 \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_4 \operatorname {BesselY}\left (-\sqrt {5}, x\right )\right )}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {\operatorname {BesselJ}\left (-\sqrt {5}, x\right ) \sqrt {5}\, c_3 +\operatorname {BesselJ}\left (-\sqrt {5}+1, x\right ) c_3 x +\sqrt {5}\, \operatorname {BesselY}\left (-\sqrt {5}, x\right ) c_4 +\operatorname {BesselY}\left (-\sqrt {5}+1, x\right ) c_4 x}{x \left (c_3 \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_4 \operatorname {BesselY}\left (-\sqrt {5}, x\right )\right )}d x}\\ &= \frac {1}{c_3 \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_4 \operatorname {BesselY}\left (-\sqrt {5}, 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 (-\sqrt {5}, x\right )+c_4 \operatorname {BesselY}\left (-\sqrt {5}, x\right )}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} \frac {y}{c_3 \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_4 \operatorname {BesselY}\left (-\sqrt {5}, x\right )}&= \int {0 \,dx} + c_5 \\ &=c_5 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{c_3 \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_4 \operatorname {BesselY}\left (-\sqrt {5}, x\right )}\) gives the final solution

\[ y = \left (c_3 \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_4 \operatorname {BesselY}\left (-\sqrt {5}, x\right )\right ) c_5 \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \left (c_3 \operatorname {BesselJ}\left (-\sqrt {5}, x\right )+c_4 \operatorname {BesselY}\left (-\sqrt {5}, x\right )\right ) c_5 \\ \end{align*}

Will add steps showing solving for IC soon.

2.45.3 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+x y^{\prime }+\left (x^{2}-5\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 (x^{2}-5\right ) y}{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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {y^{\prime }}{x}+\frac {\left (x^{2}-5\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 {1}{x}, P_{3}\left (x \right )=\frac {x^{2}-5}{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}}}=-5 \\ {} & \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 )+x y^{\prime }+\left (x^{2}-5\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}-5\right ) x^{r}+a_{1} \left (r^{2}+2 r -4\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}-5\right )+a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}-5=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\sqrt {5}, -\sqrt {5}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (r^{2}+2 r -4\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 (k^{2}+2 k r +r^{2}-5\right )+a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (\left (k +2\right )^{2}+2 \left (k +2\right ) r +r^{2}-5\right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{k^{2}+2 k r +r^{2}+4 k +4 r -1} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\sqrt {5} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{k^{2}+2 k \sqrt {5}+4+4 k +4 \sqrt {5}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\sqrt {5} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\sqrt {5}}, a_{k +2}=-\frac {a_{k}}{k^{2}+2 k \sqrt {5}+4+4 k +4 \sqrt {5}}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\sqrt {5} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{k^{2}-2 k \sqrt {5}+4+4 k -4 \sqrt {5}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\sqrt {5} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\sqrt {5}}, a_{k +2}=-\frac {a_{k}}{k^{2}-2 k \sqrt {5}+4+4 k -4 \sqrt {5}}, 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 +\sqrt {5}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\sqrt {5}}\right ), a_{k +2}=-\frac {a_{k}}{k^{2}+2 k \sqrt {5}+4+4 k +4 \sqrt {5}}, a_{1}=0, b_{k +2}=-\frac {b_{k}}{k^{2}-2 k \sqrt {5}+4+4 k -4 \sqrt {5}}, b_{1}=0\right ] \end {array} \]

2.45.4 Maple trace
Methods for second order ODEs:
 
2.45.5 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 19

dsolve(x^2*diff(diff(y(x),x),x)+x*diff(y(x),x)+(x^2-5)*y(x) = 0, 
       y(x),singsol=all)
 
\[ y = c_1 \operatorname {BesselJ}\left (\sqrt {5}, x\right )+c_2 \operatorname {BesselY}\left (\sqrt {5}, x\right ) \]
2.45.6 Mathematica DSolve solution

Solving time : 0.133 (sec)
Leaf size : 26

DSolve[{x^2*D[y[x],{x,2}]+x*D[y[x],x]+(x^2-5)*y[x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 
\[ y(x)\to c_1 \operatorname {BesselJ}\left (\sqrt {5},x\right )+c_2 \operatorname {BesselY}\left (\sqrt {5},x\right ) \]