9.25 problem 265

9.25.1 Solving as riccati ode
9.25.2 Maple step by step solution

Internal problem ID [3521]
Internal file name [OUTPUT/3014_Sunday_June_05_2022_08_49_38_AM_16724665/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 9
Problem number: 265.
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]

\[ \boxed {y^{\prime } x^{2}+a x \left (1-y x \right )-x^{2} y^{2}=-2} \]

9.25.1 Solving as riccati ode

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

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

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

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

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

\[ y = \frac {-c_{3} \left (x a -1\right ) \left (a^{2} x^{2}+2\right ) {\mathrm e}^{x a}+1}{x \left (\left (a^{2} x^{2}-2 x a +2\right ) {\mathrm e}^{x a} c_{3} +1\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-c_{3} \left (x a -1\right ) \left (a^{2} x^{2}+2\right ) {\mathrm e}^{x a}+1}{x \left (\left (a^{2} x^{2}-2 x a +2\right ) {\mathrm e}^{x a} c_{3} +1\right )} \\ \end{align*}

Verification of solutions

\[ y = \frac {-c_{3} \left (x a -1\right ) \left (a^{2} x^{2}+2\right ) {\mathrm e}^{x a}+1}{x \left (\left (a^{2} x^{2}-2 x a +2\right ) {\mathrm e}^{x a} c_{3} +1\right )} \] Verified OK.

9.25.2 Maple step by step solution

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

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) = a*(diff(y(x), x))+(a*x+2)*y(x)/x^2, 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)] 
      <- linear_1 successful 
   <- Riccati to 2nd Order successful`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 52

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

\[ y \left (x \right ) = \frac {-\left (a x -1\right ) \left (x^{2} a^{2}+2\right ) {\mathrm e}^{a x}+c_{1}}{x \left (\left (x^{2} a^{2}-2 a x +2\right ) {\mathrm e}^{a x}+c_{1} \right )} \]

Solution by Mathematica

Time used: 0.366 (sec). Leaf size: 78

DSolve[x^2 y'[x]+2+a x(1-x y[x])-x^2 y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to \frac {e^{a x} \left (-a^3 x^3+a^2 x^2-2 a x+2\right )+a^3 c_1}{x \left (e^{a x} \left (a^2 x^2-2 a x+2\right )+a^3 c_1\right )} \\ y(x)\to \frac {1}{x} \\ \end{align*}