problem Problem 1(a)

Internal problem ID [12221]
Internal ﬁle name [OUTPUT/10874_Thursday_September_21_2023_05_48_23_AM_65554067/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 4, Second and Higher Order Linear Diﬀerential Equations. Problems page 221
Problem number: Problem 1(a).
ODE order: 2.
ODE degree: 1.

2.1.1 Solving as second order bessel ode ode

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

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

Summary

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

Veriﬁcation of solutions

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

2.1.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+y x^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -2 \\ {} & {} & x^{2}\cdot y=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k -2} x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) x^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) x^{k} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 6 a_{3} x +2 a_{2}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )+a_{k -2}\right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [2 a_{2}=0, 6 a_{3}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{2}=0, a_{3}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+3 k +2\right ) a_{k +2}+a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+3 k +8\right ) a_{k +4}+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=-\frac {a_{k}}{k^{2}+7 k +12}, a_{2}=0, a_{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`

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

\[ y(x)\to c_2 \operatorname {ParabolicCylinderD}\left (-\frac {1}{2},(-1+i) x\right )+c_1 \operatorname {ParabolicCylinderD}\left (-\frac {1}{2},(1+i) x\right ) \]

2.1 problem Problem 1(a)

2.1.1 Solving as second order bessel ode ode

2.1.2 Maple step by step solution