Internal problem ID [10389]
Internal file name [OUTPUT/9337_Monday_June_06_2022_01_57_00_PM_79642548/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: 60.
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]
\[ \boxed {\left (x^{2} a +b x +c \right ) y^{\prime }-y^{2}-\left (x a +\mu \right ) y=-\lambda ^{2} x^{2}+\lambda \left (b -\mu \right ) x +c \lambda } \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-\lambda ^{2} x^{2}+a x y +b \lambda x -\lambda x \mu +c \lambda +\mu y +y^{2}}{x^{2} a +b x +c} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {\lambda ^{2} x^{2}}{x^{2} a +b x +c}+\frac {a x y}{x^{2} a +b x +c}+\frac {b \lambda x}{x^{2} a +b x +c}-\frac {\lambda x \mu }{x^{2} a +b x +c}+\frac {c \lambda }{x^{2} a +b x +c}+\frac {\mu y}{x^{2} a +b x +c}+\frac {y^{2}}{x^{2} a +b x +c} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {-\lambda ^{2} x^{2}+b \lambda x -\lambda x \mu +c \lambda }{x^{2} a +b x +c}\), \(f_1(x)=\frac {x a +\mu }{x^{2} a +b x +c}\) and \(f_2(x)=\frac {1}{x^{2} a +b x +c}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{x^{2} a +b x +c}} \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 x a +b}{\left (x^{2} a +b x +c \right )^{2}}\\ f_1 f_2 &=\frac {x a +\mu }{\left (x^{2} a +b x +c \right )^{2}}\\ f_2^2 f_0 &=\frac {-\lambda ^{2} x^{2}+b \lambda x -\lambda x \mu +c \lambda }{\left (x^{2} a +b x +c \right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{x^{2} a +b x +c}-\left (-\frac {2 x a +b}{\left (x^{2} a +b x +c \right )^{2}}+\frac {x a +\mu }{\left (x^{2} a +b x +c \right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {\left (-\lambda ^{2} x^{2}+b \lambda x -\lambda x \mu +c \lambda \right ) u \left (x \right )}{\left (x^{2} a +b x +c \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
\[ \text {Expression too large to display} \]
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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2} a +b x +c \right ) y^{\prime }-y^{2}-\left (x a +\mu \right ) y=-\lambda ^{2} x^{2}+\lambda \left (b -\mu \right ) x +c \lambda \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+\left (x a +\mu \right ) y-\lambda ^{2} x^{2}+\lambda \left (b -\mu \right ) x +c \lambda }{x^{2} a +b x +c} \end {array} \]
Maple trace Kovacic algorithm successful
`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*x+b-mu)*(diff(y(x), x))/(a*x^2+b*x+c)-lambda*(-lambda*x^2+b*x-mu*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 A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> 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 -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals <- Kovacics algorithm successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 6204
dsolve((a*x^2+b*x+c)*diff(y(x),x)=y(x)^2+(a*x+mu)*y(x)-lambda^2*x^2+lambda*(b-mu)*x+lambda*c,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 23.352 (sec). Leaf size: 433
DSolve[(a*x^2+b*x+c)*y'[x]==y[x]^2+(a*x+\[Mu])*y[x]-\[Lambda]^2*x^2+\[Lambda]*(b-\[Mu])*x+\[Lambda]*c,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {(x (a x+b)+c)^{\frac {\lambda }{a}-\frac {1}{2}} \exp \left (-\frac {(a (b-2 \mu )+2 b \lambda ) \arctan \left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{a \sqrt {4 a c-b^2}}\right ) \left (\lambda x (x (a x+b)+c)^{\frac {1}{2}-\frac {\lambda }{a}} \exp \left (\frac {(a (b-2 \mu )+2 b \lambda ) \arctan \left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{a \sqrt {4 a c-b^2}}\right ) \int _1^x\exp \left (-\frac {\frac {2 (2 b \lambda +a (b-2 \mu )) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+(a-2 \lambda ) \log (c+K[1] (b+a K[1]))}{2 a}\right )dK[1]+x \left (c_1 \lambda (x (a x+b)+c)^{\frac {1}{2}-\frac {\lambda }{a}} \exp \left (\frac {(a (b-2 \mu )+2 b \lambda ) \arctan \left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{a \sqrt {4 a c-b^2}}\right )-a x-b\right )-c\right )}{\int _1^x\exp \left (-\frac {\frac {2 (2 b \lambda +a (b-2 \mu )) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+(a-2 \lambda ) \log (c+K[1] (b+a K[1]))}{2 a}\right )dK[1]+c_1} \\ y(x)\to \lambda x \\ \end{align*}