Internal problem ID [10487]
Internal file name [OUTPUT/9435_Monday_June_06_2022_02_32_20_PM_19258479/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. subsection 1.2.5-2
Problem number: 13.
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]
\[ \boxed {y^{\prime }+\left (1+n \right ) x^{n} y^{2}-a \,x^{1+n} \ln \left (x \right )^{m} y=-a \ln \left (x \right )^{m}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -x^{n} y^{2} n +a \,x^{1+n} \ln \left (x \right )^{m} y -x^{n} y^{2}-a \ln \left (x \right )^{m} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -x^{n} y^{2} n +a x \,x^{n} \ln \left (x \right )^{m} y -x^{n} y^{2}-a \ln \left (x \right )^{m} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a \ln \left (x \right )^{m}\), \(f_1(x)=a \,x^{1+n} \ln \left (x \right )^{m}\) and \(f_2(x)=-n \,x^{n}-x^{n}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\left (-n \,x^{n}-x^{n}\right ) 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 {n^{2} x^{n}}{x}-\frac {x^{n} n}{x}\\ f_1 f_2 &=a \,x^{1+n} \ln \left (x \right )^{m} \left (-n \,x^{n}-x^{n}\right )\\ f_2^2 f_0 &=-\left (-n \,x^{n}-x^{n}\right )^{2} a \ln \left (x \right )^{m} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \left (-n \,x^{n}-x^{n}\right ) u^{\prime \prime }\left (x \right )-\left (-\frac {n^{2} x^{n}}{x}-\frac {x^{n} n}{x}+a \,x^{1+n} \ln \left (x \right )^{m} \left (-n \,x^{n}-x^{n}\right )\right ) u^{\prime }\left (x \right )-\left (-n \,x^{n}-x^{n}\right )^{2} a \ln \left (x \right )^{m} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{1+n} \left (\left (\int x^{-2 n -2} {\mathrm e}^{\int \left (a \,x^{1+n} \ln \left (x \right )^{m}+\frac {n}{x}\right )d x}d x \right ) c_{2} +c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = c_{2} x^{-1-n} {\mathrm e}^{\int \frac {\ln \left (x \right )^{m} x^{2+n} a +n}{x}d x}+x^{n} \left (1+n \right ) \left (\left (\int {\mathrm e}^{\int \frac {\ln \left (x \right )^{m} x^{2+n} a +n}{x}d x} x^{-2 n -2}d x \right ) c_{2} +c_{1} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {\left (c_{2} x^{-1-n} {\mathrm e}^{\int \frac {\ln \left (x \right )^{m} x^{2+n} a +n}{x}d x}+x^{n} \left (1+n \right ) \left (\left (\int {\mathrm e}^{\int \frac {\ln \left (x \right )^{m} x^{2+n} a +n}{x}d x} x^{-2 n -2}d x \right ) c_{2} +c_{1} \right )\right ) x^{-1-n}}{\left (-n \,x^{n}-x^{n}\right ) \left (\left (\int x^{-2 n -2} {\mathrm e}^{\int \left (a \,x^{1+n} \ln \left (x \right )^{m}+\frac {n}{x}\right )d x}d x \right ) c_{2} +c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\int x^{-2 n -2} {\mathrm e}^{\int \left (a \,x^{1+n} \ln \left (x \right )^{m}+\frac {n}{x}\right )d x}d x +c_{3} \right ) \left (1+n \right ) x^{-1-n}+x^{-2-3 n} {\mathrm e}^{\int \left (a \,x^{1+n} \ln \left (x \right )^{m}+\frac {n}{x}\right )d x}}{\left (1+n \right ) \left (\int {\mathrm e}^{\int \frac {\ln \left (x \right )^{m} x^{2+n} a +n}{x}d x} x^{-2 n -2}d x +c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\int x^{-2 n -2} {\mathrm e}^{\int \left (a \,x^{1+n} \ln \left (x \right )^{m}+\frac {n}{x}\right )d x}d x +c_{3} \right ) \left (1+n \right ) x^{-1-n}+x^{-2-3 n} {\mathrm e}^{\int \left (a \,x^{1+n} \ln \left (x \right )^{m}+\frac {n}{x}\right )d x}}{\left (1+n \right ) \left (\int {\mathrm e}^{\int \frac {\ln \left (x \right )^{m} x^{2+n} a +n}{x}d x} x^{-2 n -2}d x +c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\int x^{-2 n -2} {\mathrm e}^{\int \left (a \,x^{1+n} \ln \left (x \right )^{m}+\frac {n}{x}\right )d x}d x +c_{3} \right ) \left (1+n \right ) x^{-1-n}+x^{-2-3 n} {\mathrm e}^{\int \left (a \,x^{1+n} \ln \left (x \right )^{m}+\frac {n}{x}\right )d x}}{\left (1+n \right ) \left (\int {\mathrm e}^{\int \frac {\ln \left (x \right )^{m} x^{2+n} a +n}{x}d x} x^{-2 n -2}d x +c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\left (1+n \right ) x^{n} y^{2}-a \,x^{1+n} \ln \left (x \right )^{m} y=-a \ln \left (x \right )^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\left (1+n \right ) x^{n} y^{2}+a \,x^{1+n} \ln \left (x \right )^{m} y-a \ln \left (x \right )^{m} \end {array} \]
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) = (a*x^(n+1)*ln(x)^m*x+n)*(diff(y(x), x))/x-x^n*(n+1)*a*ln(x)^m*y(x), y( 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 -> 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 trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> 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 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 <- 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 -> 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 trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> 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 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 <- 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)-((-x^n*n-x^n)*y(x)^2+y(x)+a*x^(n+1)*ln(x)^m*y(x)*x-x^2*a*ln(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
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 184
dsolve(diff(y(x),x)=-(n+1)*x^n*y(x)^2+a*x^(n+1)*(ln(x))^m*y(x)-a*(ln(x))^m,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{-n -1} \left (x^{n +1} {\mathrm e}^{\int \frac {a \,x^{n +1} \ln \left (x \right )^{m} x -2 n -2}{x}d x}+\left (\int x^{n} {\mathrm e}^{a \left (\int x^{n +1} \ln \left (x \right )^{m}d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (n +1\right )}d x \right ) n +\int x^{n} {\mathrm e}^{a \left (\int x^{n +1} \ln \left (x \right )^{m}d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (n +1\right )}d x -c_{1} \right )}{\left (\int x^{n} {\mathrm e}^{a \left (\int x^{n +1} \ln \left (x \right )^{m}d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (n +1\right )}d x \right ) n +\int x^{n} {\mathrm e}^{a \left (\int x^{n +1} \ln \left (x \right )^{m}d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (n +1\right )}d x -c_{1}} \]
✓ Solution by Mathematica
Time used: 5.364 (sec). Leaf size: 311
DSolve[y'[x]==-(n+1)*x^n*y[x]^2+a*x^(n+1)*(Log[x])^m*y[x]-a*(Log[x])^m,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^{-2 (n+1)} \left (c_1 (n+1) x^{n+1} \int _1^x\exp \left (\frac {a \Gamma (m+1,-((n+2) \log (K[1]))) \log ^m(K[1]) (-((n+2) \log (K[1])))^{-m}}{n+2}-(n+2) \log (K[1])\right )dK[1]+c_1 \exp \left (\frac {a \log ^m(x) (-((n+2) \log (x)))^{-m} \Gamma (m+1,-((n+2) \log (x)))}{n+2}\right )+(n+1) x^{n+1}\right )}{(n+1) \left (1+c_1 \int _1^x\exp \left (\frac {a \Gamma (m+1,-((n+2) \log (K[1]))) \log ^m(K[1]) (-((n+2) \log (K[1])))^{-m}}{n+2}-(n+2) \log (K[1])\right )dK[1]\right )} \\ y(x)\to \frac {x^{-2 (n+1)} \left (\frac {\exp \left (\frac {a \log ^m(x) (-((n+2) \log (x)))^{-m} \Gamma (m+1,-((n+2) \log (x)))}{n+2}\right )}{\int _1^x\exp \left (\frac {a \Gamma (m+1,-((n+2) \log (K[1]))) \log ^m(K[1]) (-((n+2) \log (K[1])))^{-m}}{n+2}-(n+2) \log (K[1])\right )dK[1]}+(n+1) x^{n+1}\right )}{n+1} \\ \end{align*}