6.17 problem 163

Internal problem ID [3419]
Internal file name [OUTPUT/2912_Sunday_June_05_2022_08_46_56_AM_66503190/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 6
Problem number: 163.
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]

\[ \boxed {x y^{\prime }+y^{2}=-x^{2}} \]

6.17.1 Solving as riccati ode

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

This is a Riccati ODE. Comparing the ODE to solve \[ y' = -x -\frac {y^{2}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-x\), \(f_1(x)=0\) and \(f_2(x)=-\frac {1}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {u}{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}{x^{2}}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=-\frac {1}{x} \end {align*}

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

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

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

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

Summary

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

Figure 229: Slope field plot

Verification of solutions

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

6.17.2 Maple step by step solution

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

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 Riccati to 2nd Order 
   -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(diff(y(x), x))/x-y(x), y(x)`      *** Sublevel 2 *** 
      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 
   <- Riccati to 2nd Order successful`
 

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 27

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

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

✓ Solution by Mathematica

Time used: 0.163 (sec). Leaf size: 45

DSolve[x y'[x]+x^2+y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -\frac {x (\operatorname {BesselY}(1,x)+c_1 \operatorname {BesselJ}(1,x))}{\operatorname {BesselY}(0,x)+c_1 \operatorname {BesselJ}(0,x)} \\ y(x)\to -\frac {x \operatorname {BesselJ}(1,x)}{\operatorname {BesselJ}(0,x)} \\ \end{align*}