Internal problem ID [10374]
Internal file name [OUTPUT/9322_Monday_June_06_2022_01_51_26_PM_43318158/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: 45.
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]
\[ \boxed {\left (a_{2} x +b_{2} \right ) \left (y^{\prime }+\lambda y^{2}\right )+\left (a_{1} x +b_{1} \right ) y=-a_{0} x -b_{0}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {y^{2} a_{2} \lambda x +y^{2} b_{2} \lambda +y a_{1} x +a_{0} x +y b_{1} +b_{0}}{a_{2} x +b_{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {y^{2} a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {y^{2} b_{2} \lambda }{a_{2} x +b_{2}}-\frac {y a_{1} x}{a_{2} x +b_{2}}-\frac {a_{0} x}{a_{2} x +b_{2}}-\frac {y b_{1}}{a_{2} x +b_{2}}-\frac {b_{0}}{a_{2} x +b_{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {a_{0} x +b_{0}}{a_{2} x +b_{2}}\), \(f_1(x)=-\frac {a_{1} x +b_{1}}{a_{2} x +b_{2}}\) and \(f_2(x)=-\frac {a_{2} \lambda x +b_{2} \lambda }{a_{2} x +b_{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {\left (a_{2} \lambda x +b_{2} \lambda \right ) u}{a_{2} x +b_{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 {a_{2} \lambda }{a_{2} x +b_{2}}+\frac {\left (a_{2} \lambda x +b_{2} \lambda \right ) a_{2}}{\left (a_{2} x +b_{2} \right )^{2}}\\ f_1 f_2 &=\frac {\left (a_{1} x +b_{1} \right ) \left (a_{2} \lambda x +b_{2} \lambda \right )}{\left (a_{2} x +b_{2} \right )^{2}}\\ f_2^2 f_0 &=-\frac {\left (a_{2} \lambda x +b_{2} \lambda \right )^{2} \left (a_{0} x +b_{0} \right )}{\left (a_{2} x +b_{2} \right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {\left (a_{2} \lambda x +b_{2} \lambda \right ) u^{\prime \prime }\left (x \right )}{a_{2} x +b_{2}}-\left (-\frac {a_{2} \lambda }{a_{2} x +b_{2}}+\frac {\left (a_{2} \lambda x +b_{2} \lambda \right ) a_{2}}{\left (a_{2} x +b_{2} \right )^{2}}+\frac {\left (a_{1} x +b_{1} \right ) \left (a_{2} \lambda x +b_{2} \lambda \right )}{\left (a_{2} x +b_{2} \right )^{2}}\right ) u^{\prime }\left (x \right )-\frac {\left (a_{2} \lambda x +b_{2} \lambda \right )^{2} \left (a_{0} x +b_{0} \right ) u \left (x \right )}{\left (a_{2} x +b_{2} \right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (a_{2} x +b_{2} \right )^{\frac {b_{2} a_{1} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}} {\mathrm e}^{-\frac {\left (\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}+a_{1} \right ) x}{2 a_{2}}} \left (\operatorname {KummerM}\left (\frac {\left (b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1} \right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}-2 a_{2}^{2} b_{0} \lambda +\left (2 a_{0} \lambda b_{2} +a_{1} b_{1} \right ) a_{2} -a_{1}^{2} b_{2}}{2 \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{2}^{2}}, \frac {b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\right ) c_{1} +\operatorname {KummerU}\left (\frac {\left (b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1} \right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}-2 a_{2}^{2} b_{0} \lambda +\left (2 a_{0} \lambda b_{2} +a_{1} b_{1} \right ) a_{2} -a_{1}^{2} b_{2}}{2 \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{2}^{2}}, \frac {b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {2 \,{\mathrm e}^{-\frac {\left (\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}+a_{1} \right ) x}{2 a_{2}}} \left (c_{1} \left (\frac {\left (a_{1} x +b_{1} \right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}}{4}+\lambda \left (a_{0} x +\frac {b_{0}}{2}\right ) a_{2} -\frac {x \,a_{1}^{2}}{4}+\frac {a_{0} \lambda b_{2}}{2}-\frac {a_{1} b_{1}}{4}\right ) a_{2} \operatorname {KummerM}\left (\frac {\left (b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1} \right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}-2 a_{2}^{2} b_{0} \lambda +\left (2 a_{0} \lambda b_{2} +a_{1} b_{1} \right ) a_{2} -a_{1}^{2} b_{2}}{2 \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{2}^{2}}, \frac {b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\right )+\left (\frac {\left (a_{1} x +b_{1} \right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}}{4}+\lambda \left (a_{0} x +\frac {b_{0}}{2}\right ) a_{2} -\frac {x \,a_{1}^{2}}{4}+\frac {a_{0} \lambda b_{2}}{2}-\frac {a_{1} b_{1}}{4}\right ) c_{2} a_{2} \operatorname {KummerU}\left (\frac {\left (b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1} \right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}-2 a_{2}^{2} b_{0} \lambda +\left (2 a_{0} \lambda b_{2} +a_{1} b_{1} \right ) a_{2} -a_{1}^{2} b_{2}}{2 \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{2}^{2}}, \frac {b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\right )+\frac {c_{1} \left (\left (-\frac {1}{2} b_{2} a_{1} -a_{2}^{2}+\frac {1}{2} a_{2} b_{1} \right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}-a_{2}^{2} b_{0} \lambda +\left (a_{0} \lambda b_{2} +\frac {a_{1} b_{1}}{2}\right ) a_{2} -\frac {a_{1}^{2} b_{2}}{2}\right ) \operatorname {KummerM}\left (\frac {2 a_{0} a_{2} \lambda b_{2} -2 a_{2}^{2} b_{0} \lambda +\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{1} b_{2} -\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{2} b_{1} -a_{1}^{2} b_{2} +a_{1} a_{2} b_{1}}{2 a_{2}^{2} \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}}, \frac {b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\right )}{2}+\frac {\operatorname {KummerU}\left (\frac {2 a_{0} a_{2} \lambda b_{2} -2 a_{2}^{2} b_{0} \lambda +\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{1} b_{2} -\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{2} b_{1} -a_{1}^{2} b_{2} +a_{1} a_{2} b_{1}}{2 a_{2}^{2} \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}}, \frac {b_{2} a_{1} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, c_{2} a_{2}^{2}}{2}\right ) \left (a_{2} x +b_{2} \right )^{\frac {b_{2} a_{1} -a_{2} b_{1}}{a_{2}^{2}}}}{\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{2}} \] 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
\[ \text {Expression too large to display} \]
The solution(s) found are the following \begin{align*} \tag{1} \text {Expression too large to display} \\ \end{align*}
Verification of solutions
\[ \text {Expression too large to display} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a_{2} x +b_{2} \right ) \left (y^{\prime }+\lambda y^{2}\right )+\left (a_{1} x +b_{1} \right ) y=-a_{0} x -b_{0} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y^{2} a_{2} \lambda x +y^{2} b_{2} \lambda +y a_{1} x +y b_{1} +a_{0} x +b_{0}}{a_{2} x +b_{2}} \end {array} \]
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+b__1)*(diff(y(x), x))/(a__2*x+b__2)-lambda*(a__0*x+b__0)*y(x) 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 -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 827
dsolve((a__2*x+b__2)*(diff(y(x),x)+lambda*y(x)^2)+(a__1*x+b__1)*y(x)+a__0*x+b__0=0,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 3.165 (sec). Leaf size: 1418
DSolve[(a2*x+b2)*(y'[x]+\[Lambda]*y[x]^2)+(a1*x+b1)*y[x]+a0*x+b0==0,y[x],x,IncludeSingularSolutions -> True]
Too large to display