Internal problem ID [10361]
Internal file name [OUTPUT/9309_Monday_June_06_2022_01_51_03_PM_81658841/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: 32.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime }+a n \,x^{-1+n} y^{2}-c \,x^{m} \left (x^{n} a +b \right ) y=-c \,x^{m}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -a n \,x^{-1+n} y^{2}+x^{m} x^{n} a c y +x^{m} b c y -c \,x^{m} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {a \,x^{n} n \,y^{2}}{x}+x^{m} x^{n} a c y +x^{m} b c y -c \,x^{m} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-c \,x^{m}\), \(f_1(x)=x^{m} x^{n} a c +x^{m} b c\) and \(f_2(x)=-a n \,x^{-1+n}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-a n \,x^{-1+n} 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' &=-\frac {a n \,x^{-1+n} \left (-1+n \right )}{x}\\ f_1 f_2 &=-\left (x^{m} x^{n} a c +x^{m} b c \right ) a n \,x^{-1+n}\\ f_2^2 f_0 &=-a^{2} n^{2} x^{2 n -2} c \,x^{m} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -a n \,x^{-1+n} u^{\prime \prime }\left (x \right )-\left (-\frac {a n \,x^{-1+n} \left (-1+n \right )}{x}-\left (x^{m} x^{n} a c +x^{m} b c \right ) a n \,x^{-1+n}\right ) u^{\prime }\left (x \right )-a^{2} n^{2} x^{2 n -2} c \,x^{m} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\left (\int \frac {x^{-1+n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{\left (x^{n} a +b \right )^{2}}d x \right ) c_{2} +c_{1} \right ) \left (x^{n} a +b \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {x^{n} \left (a n c_{2} \left (x^{n} a +b \right ) \left (\int \frac {x^{n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{x \left (x^{n} a +b \right )^{2}}d x \right )+{\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}} c_{2} +a n c_{1} \left (x^{n} a +b \right )\right )}{x \left (x^{n} a +b \right )} \] Using the above in (1) gives the solution \[ y = \frac {x^{n} \left (a n c_{2} \left (x^{n} a +b \right ) \left (\int \frac {x^{n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{x \left (x^{n} a +b \right )^{2}}d x \right )+{\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}} c_{2} +a n c_{1} \left (x^{n} a +b \right )\right ) x^{1-n}}{x \left (x^{n} a +b \right )^{2} a n \left (\left (\int \frac {x^{-1+n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{\left (x^{n} a +b \right )^{2}}d x \right ) c_{2} +c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {a n \left (x^{n} a +b \right ) \left (\int \frac {x^{n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{x \left (x^{n} a +b \right )^{2}}d x \right )+a^{2} n c_{3} x^{n}+a n c_{3} b +{\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{a n \left (x^{n} a +b \right )^{2} \left (\int \frac {x^{n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{x \left (x^{n} a +b \right )^{2}}d x +c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {a n \left (x^{n} a +b \right ) \left (\int \frac {x^{n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{x \left (x^{n} a +b \right )^{2}}d x \right )+a^{2} n c_{3} x^{n}+a n c_{3} b +{\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{a n \left (x^{n} a +b \right )^{2} \left (\int \frac {x^{n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{x \left (x^{n} a +b \right )^{2}}d x +c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {a n \left (x^{n} a +b \right ) \left (\int \frac {x^{n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{x \left (x^{n} a +b \right )^{2}}d x \right )+a^{2} n c_{3} x^{n}+a n c_{3} b +{\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{a n \left (x^{n} a +b \right )^{2} \left (\int \frac {x^{n} {\mathrm e}^{\frac {\left (a \left (m +1\right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m} x}{\left (m +1\right ) \left (1+m +n \right )}}}{x \left (x^{n} a +b \right )^{2}}d x +c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+a n \,x^{-1+n} y^{2}-c \,x^{m} \left (x^{n} a +b \right ) y=-c \,x^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-a n \,x^{-1+n} y^{2}+c \,x^{m} \left (x^{n} a +b \right ) y-c \,x^{m} \end {array} \]
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_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (x*x^m*b*c+x^(n+m)*a*c*x+n-1)*(diff(y(x), x))/x-a*n*x^(n-1)*c*x^m*y(x) Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying an equivalence, under non-integer power transformations, to LODEs admitting Liouvillian solutions. -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying an equivalence, under non-integer power transformations, to LODEs admitting Liouvillian solutions. -> 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 -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> Trying a change of variables to reduce to Bernoulli -> Calling odsolve with the ODE`, diff(y(x), x)-(-a*n*x^(n-1)*y(x)^2+y(x)+(x^(n+m)*a*c+x^m*b*c)*y(x)*x-x^2*c*x^m)/x, y(x), explic 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_symmetries trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 2 `, `-> Computing symmetries using: way = 6`[0, exp(-b*c/(1+m)*x*x^m-a*c/(1+m+n)*x*x^m*x^n+2*ln(a*x^n+b))*(y-x^n/x/(x^(n-1)
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 199
dsolve(diff(y(x),x)=-a*n*x^(n-1)*y(x)^2+c*x^m*(a*x^n+b)*y(x)-c*x^m,y(x), singsol=all)
\[ y \left (x \right ) = \frac {a n \left (a \,x^{n}+b \right ) \left (\int \frac {x^{n -1} {\mathrm e}^{\frac {c \left (a \left (1+m \right ) x^{1+m +n}+b \,x^{1+m} \left (1+m +n \right )\right )}{\left (1+m \right ) \left (1+m +n \right )}}}{\left (a \,x^{n}+b \right )^{2}}d x \right )-x^{n} c_{1} a -c_{1} b +{\mathrm e}^{\frac {c \left (a \left (1+m \right ) x^{1+m +n}+b \,x^{1+m} \left (1+m +n \right )\right )}{\left (1+m \right ) \left (1+m +n \right )}}}{\left (a \left (\int \frac {x^{n -1} {\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (1+m +n \right )\right ) c \,x^{m}}{\left (1+m \right ) \left (1+m +n \right )}}}{\left (a \,x^{n}+b \right )^{2}}d x \right ) n -c_{1} \right ) \left (a^{2} x^{2 n}+2 x^{n} a b +b^{2}\right )} \]
✓ Solution by Mathematica
Time used: 8.659 (sec). Leaf size: 304
DSolve[y'[x]==-a*n*x^(n-1)*y[x]^2+c*x^m*(a*x^n+b)*y[x]-c*x^m,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {a c_1 n \left (a x^n+b\right ) \int _1^x\frac {\exp \left (c K[1]^{m+1} \left (\frac {a K[1]^n}{m+n+1}+\frac {b}{m+1}\right )\right ) K[1]^{n-1}}{\left (a K[1]^n+b\right )^2}dK[1]+a^2 n x^n+c_1 e^{c x^{m+1} \left (\frac {a x^n}{m+n+1}+\frac {b}{m+1}\right )}+a b n}{a n \left (a x^n+b\right )^2 \left (1+c_1 \int _1^x\frac {\exp \left (c K[1]^{m+1} \left (\frac {a K[1]^n}{m+n+1}+\frac {b}{m+1}\right )\right ) K[1]^{n-1}}{\left (a K[1]^n+b\right )^2}dK[1]\right )} \\ y(x)\to \frac {\frac {e^{c x^{m+1} \left (\frac {a x^n}{m+n+1}+\frac {b}{m+1}\right )}}{a n \int _1^x\frac {\exp \left (c K[1]^{m+1} \left (\frac {a K[1]^n}{m+n+1}+\frac {b}{m+1}\right )\right ) K[1]^{n-1}}{\left (a K[1]^n+b\right )^2}dK[1]}+a x^n+b}{\left (a x^n+b\right )^2} \\ \end{align*}