Internal problem ID [10357]
Internal file name [OUTPUT/9305_Monday_June_06_2022_01_50_51_PM_42556060/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: 28.
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]
\[ \boxed {y^{\prime }-y^{2}-a \,x^{n} y=-b a \,x^{n}-b^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+x^{n} a y -b a \,x^{n}-b^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+x^{n} a y -b a \,x^{n}-b^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-b a \,x^{n}-b^{2}\), \(f_1(x)=x^{n} a\) 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 &=x^{n} a\\ f_2^2 f_0 &=-b a \,x^{n}-b^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-x^{n} a u^{\prime }\left (x \right )+\left (-b a \,x^{n}-b^{2}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\int \frac {-\left (\int {\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}d x \right ) b +c_{1} b +{\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}}{\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x -c_{1}}d x} c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {c_{2} \left (-\left (\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x \right ) b +c_{1} b +{\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}\right ) {\mathrm e}^{\int \frac {-\left (\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x \right ) b +c_{1} b +{\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}}{\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x -c_{1}}d x}}{-\left (\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x \right )+c_{1}} \] Using the above in (1) gives the solution \[ y = \frac {\left (-\left (\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x \right ) b +c_{1} b +{\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}\right ) {\mathrm e}^{\int \frac {-\left (\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x \right ) b +c_{1} b +{\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}}{\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x -c_{1}}d x} {\mathrm e}^{\int -\frac {\int -b \,{\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}d x +{\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}+c_{1} b}{\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x -c_{1}}d x}}{-\left (\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x \right )+c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\int {\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}d x -c_{3} \right ) b -{\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}}{\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x -c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\int {\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}d x -c_{3} \right ) b -{\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}}{\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x -c_{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\int {\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}d x -c_{3} \right ) b -{\mathrm e}^{\frac {a \,x^{1+n}+2 b x \left (1+n \right )}{1+n}}}{\int {\mathrm e}^{\frac {\left (x^{n} a +2 b \left (1+n \right )\right ) x}{1+n}}d x -c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-a \,x^{n} y=-b a \,x^{n}-b^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a \,x^{n} y-b a \,x^{n}-b^{2} \end {array} \]
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_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (diff(y(x), x))*x^n*a+(b*x^n*a+b^2)*y(x), y(x)` *** Sublevel 2 ** 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 -> 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 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)-(y(x)^2+y(x)+a*x^n*y(x)*x+x^2*(-b*x^n*a-b^2))/x, y(x), explicit` *** Subleve 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)] <- symmetry pattern of the form [0, F(x)*G(y)] successful <- Riccati with symmetry pattern of the form [0,F(x)*G(y)] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 74
dsolve(diff(y(x),x)=y(x)^2+a*x^n*y(x)-a*b*x^n-b^2,y(x), singsol=all)
\[ \frac {\left (b -y \left (x \right )\right ) \left (\int _{}^{x}{\mathrm e}^{\frac {\left (\textit {\_a}^{n} a +2 b \left (n +1\right )\right ) \textit {\_a}}{n +1}}d \textit {\_a} \right )+c_{1} b -c_{1} y \left (x \right )-{\mathrm e}^{\frac {\left (a \,x^{n}+2 b \left (n +1\right )\right ) x}{n +1}}}{b -y \left (x \right )} = 0 \]
✓ Solution by Mathematica
Time used: 1.948 (sec). Leaf size: 195
DSolve[y'[x]==y[x]^2+a*x^n*y[x]-a*b*x^n-b^2,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (\frac {e^{\frac {a x^{n+1}}{n+1}+2 b x}}{a n (K[2]-b)^2}-\int _1^x\left (\frac {e^{\frac {a K[1]^{n+1}}{n+1}+2 b K[1]} \left (a K[1]^n+b+K[2]\right )}{a n (b-K[2])^2}+\frac {e^{\frac {a K[1]^{n+1}}{n+1}+2 b K[1]}}{a n (b-K[2])}\right )dK[1]\right )dK[2]+\int _1^x\frac {e^{\frac {a K[1]^{n+1}}{n+1}+2 b K[1]} \left (a K[1]^n+b+y(x)\right )}{a n (b-y(x))}dK[1]=c_1,y(x)\right ] \]