8.32 problem 237

Internal problem ID [3493]
Internal file name [OUTPUT/2986_Sunday_June_05_2022_08_48_53_AM_361749/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 8
Problem number: 237.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "riccati"

Maple gives the following as the ode type

[_rational, _Riccati]

\[ \boxed {2 x y^{\prime }-4 i x y-y^{2}=-1} \]

8.32.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {4 i x y +y^{2}-1}{2 x} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = 2 i y +\frac {y^{2}}{2 x}-\frac {1}{2 x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {1}{2 x}\), \(f_1(x)=2 i\) and \(f_2(x)=\frac {1}{2 x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{2 x}} \tag {1} \end {align*}

Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}

But \begin {align*} f_2' &=-\frac {1}{2 x^{2}}\\ f_1 f_2 &=\frac {i}{x}\\ f_2^2 f_0 &=-\frac {1}{8 x^{3}} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{2 x}-\left (\frac {i}{x}-\frac {1}{2 x^{2}}\right ) u^{\prime }\left (x \right )-\frac {u \left (x \right )}{8 x^{3}} &=0 \end {align*}

Solving the above ODE (this ode solved using Maple, not this program), gives

\[ u \left (x \right ) = \left (-c_{2} \operatorname {BesselK}\left (1, -i x \right )+c_{2} \operatorname {BesselK}\left (0, -i x \right )+c_{1} \left (i \operatorname {BesselJ}\left (1, x\right )-\operatorname {BesselJ}\left (0, x\right )\right )\right ) \sqrt {x}\, {\mathrm e}^{i x} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {{\mathrm e}^{i x} \left (-c_{2} \operatorname {BesselK}\left (1, -i x \right )-c_{2} \operatorname {BesselK}\left (0, -i x \right )+\left (i \operatorname {BesselJ}\left (1, x\right )+\operatorname {BesselJ}\left (0, x\right )\right ) c_{1} \right )}{2 \sqrt {x}} \] Using the above in (1) gives the solution \[ y = \frac {-c_{2} \operatorname {BesselK}\left (1, -i x \right )-c_{2} \operatorname {BesselK}\left (0, -i x \right )+\left (i \operatorname {BesselJ}\left (1, x\right )+\operatorname {BesselJ}\left (0, x\right )\right ) c_{1}}{-c_{2} \operatorname {BesselK}\left (1, -i x \right )+c_{2} \operatorname {BesselK}\left (0, -i x \right )+c_{1} \left (i \operatorname {BesselJ}\left (1, x\right )-\operatorname {BesselJ}\left (0, x\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = \frac {i \operatorname {BesselJ}\left (1, x\right ) c_{3} +\operatorname {BesselJ}\left (0, x\right ) c_{3} -\operatorname {BesselK}\left (0, -i x \right )-\operatorname {BesselK}\left (1, -i x \right )}{i \operatorname {BesselJ}\left (1, x\right ) c_{3} -\operatorname {BesselJ}\left (0, x\right ) c_{3} +\operatorname {BesselK}\left (0, -i x \right )-\operatorname {BesselK}\left (1, -i x \right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {i \operatorname {BesselJ}\left (1, x\right ) c_{3} +\operatorname {BesselJ}\left (0, x\right ) c_{3} -\operatorname {BesselK}\left (0, -i x \right )-\operatorname {BesselK}\left (1, -i x \right )}{i \operatorname {BesselJ}\left (1, x\right ) c_{3} -\operatorname {BesselJ}\left (0, x\right ) c_{3} +\operatorname {BesselK}\left (0, -i x \right )-\operatorname {BesselK}\left (1, -i x \right )} \\ \end{align*}

Figure 332: Slope field plot

Verification of solutions

\[ y = \frac {i \operatorname {BesselJ}\left (1, x\right ) c_{3} +\operatorname {BesselJ}\left (0, x\right ) c_{3} -\operatorname {BesselK}\left (0, -i x \right )-\operatorname {BesselK}\left (1, -i x \right )}{i \operatorname {BesselJ}\left (1, x\right ) c_{3} -\operatorname {BesselJ}\left (0, x\right ) c_{3} +\operatorname {BesselK}\left (0, -i x \right )-\operatorname {BesselK}\left (1, -i x \right )} \] Verified OK.

8.32.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x y^{\prime }-4 \,\mathrm {I} x y-y^{2}=-1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-1+4 \,\mathrm {I} x y+y^{2}}{2 x} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
differential order: 1; looking for linear symmetries 
trying exact 
Looking for potential symmetries 
trying Riccati 
trying Riccati sub-methods: 
      -> Trying a Liouvillian solution using Kovacics algorithm 
      <- No Liouvillian solutions exists 
        -> searching for a solution in terms of Whittaker functions 
        <- solution in terms of Whittaker functions successful 
   <- Abel AIR successful: ODE belongs to the 0F1 1-parameter (Bessel type) class`
 

✓ Solution by Maple

Time used: 0.031 (sec). Leaf size: 64

dsolve(2*x*diff(y(x),x)+1 = 4*I*x*y(x)+y(x)^2,y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {i \operatorname {BesselJ}\left (1, x\right )-\operatorname {BesselK}\left (1, i x \right ) c_{1} +\operatorname {BesselK}\left (0, i x \right ) c_{1} +\operatorname {BesselJ}\left (0, x\right )}{i \operatorname {BesselJ}\left (1, x\right )-\operatorname {BesselK}\left (1, i x \right ) c_{1} -\operatorname {BesselK}\left (0, i x \right ) c_{1} -\operatorname {BesselJ}\left (0, x\right )} \]

✓ Solution by Mathematica

Time used: 0.538 (sec). Leaf size: 202

DSolve[2 x y'[x]+1==4 I x y[x]+y[x]^2,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to \frac {(1-i) c_1 e^{i x} \sqrt {x} ((x-i) \operatorname {BesselJ}(0,x)-\operatorname {BesselJ}(1,x)+x \operatorname {BesselJ}(2,x))-4 i x G_{1,2}^{2,0}\left (-2 i x\left | \begin {array}{c} -1 \\ -\frac {3}{2},-\frac {1}{2} \\ \end {array} \right .\right )}{G_{1,2}^{2,0}\left (-2 i x\left | \begin {array}{c} 1 \\ -\frac {1}{2},\frac {1}{2} \\ \end {array} \right .\right )+(1+i) c_1 e^{i x} \sqrt {x} (\operatorname {BesselJ}(0,x)-i \operatorname {BesselJ}(1,x))} \\ y(x)\to -\frac {i ((x-i) \operatorname {BesselJ}(0,x)-\operatorname {BesselJ}(1,x)+x \operatorname {BesselJ}(2,x))}{\operatorname {BesselJ}(0,x)-i \operatorname {BesselJ}(1,x)} \\ y(x)\to -\frac {i ((x-i) \operatorname {BesselJ}(0,x)-\operatorname {BesselJ}(1,x)+x \operatorname {BesselJ}(2,x))}{\operatorname {BesselJ}(0,x)-i \operatorname {BesselJ}(1,x)} \\ \end{align*}