2.62 problem 62

2.62.1 Solving as riccati ode
2.62.2 Maple step by step solution

Internal problem ID [10391]
Internal file name [OUTPUT/9339_Monday_June_06_2022_02_08_40_PM_75216398/index.tex]

Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section: Chapter 1, section 1.2. Riccati Equation. 1.2.2. Equations Containing Power Functions
Problem number: 62.
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]

\[ \boxed {\left (a_{2} x^{2}+b_{2} x +c_{2} \right ) y^{\prime }-y^{2}-\left (a_{1} x +b_{1} \right ) y=a_{0} x^{2}+b_{0} x +c_{0}} \]

2.62.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a_{0} x^{2}+a_{1} x y +b_{0} x +b_{1} y +y^{2}+c_{0}}{a_{2} x^{2}+b_{2} x +c_{2}} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a_{0} x^{2}}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {a_{1} x y}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {b_{0} x}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {b_{1} y}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {y^{2}}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {c_{0}}{a_{2} x^{2}+b_{2} x +c_{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a_{0} x^{2}+b_{0} x +c_{0}}{a_{2} x^{2}+b_{2} x +c_{2}}\), \(f_1(x)=\frac {a_{1} x +b_{1}}{a_{2} x^{2}+b_{2} x +c_{2}}\) and \(f_2(x)=\frac {1}{a_{2} x^{2}+b_{2} x +c_{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{a_{2} x^{2}+b_{2} x +c_{2}}} \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 {2 a_{2} x +b_{2}}{\left (a_{2} x^{2}+b_{2} x +c_{2} \right )^{2}}\\ f_1 f_2 &=\frac {a_{1} x +b_{1}}{\left (a_{2} x^{2}+b_{2} x +c_{2} \right )^{2}}\\ f_2^2 f_0 &=\frac {a_{0} x^{2}+b_{0} x +c_{0}}{\left (a_{2} x^{2}+b_{2} x +c_{2} \right )^{3}} \end {align*}

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

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

\[ \text {Expression too large to display} \] The above shows that \[ \text {Expression too large to display} \] Using the above in (1) gives the solution \[ \text {Expression too large to display} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ \text {Expression too large to display} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} \text {Expression too large to display} \\ \end{align*}

Verification of solutions

\[ \text {Expression too large to display} \] Warning, solution could not be verified

2.62.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a_{2} x^{2}+b_{2} x +c_{2} \right ) y^{\prime }-y^{2}-\left (a_{1} x +b_{1} \right ) y=a_{0} x^{2}+b_{0} x +c_{0} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+\left (a_{1} x +b_{1} \right ) y+a_{0} x^{2}+b_{0} x +c_{0}}{a_{2} x^{2}+b_{2} x +c_{2}} \end {array} \]

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__1*x-2*a__2*x+b__1-b__2)*(diff(y(x), x))/(a__2*x^2+b__2*x+c__2)-(a_ 
      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 
         -> elliptic 
         -> Legendre 
         -> Whittaker 
            -> hyper3: Equivalence to 1F1 under a power @ Moebius 
         -> hypergeometric 
            -> heuristic approach 
            -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
            <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
         <- hypergeometric successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful`
 

Solution by Maple

Time used: 0.015 (sec). Leaf size: 6462

dsolve((a__2*x^2+b__2*x+c__2)*diff(y(x),x)=y(x)^2+(a__1*x+b__1)*y(x)+a__0*x^2+b__0*x+c__0,y(x), singsol=all)
 

\[ \text {Expression too large to display} \]

Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

DSolve[(a2*x^2+b2*x+c2)*y'[x]==y[x]^2+(a1*x+b1)*y[x]+a0*x^2+b0*x+c0,y[x],x,IncludeSingularSolutions -> True]
 

Timed out