Internal problem ID [10620]
Internal file name [OUTPUT/9568_Monday_June_06_2022_03_09_45_PM_32821835/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 1.2.8-1. Equations containing
arbitrary functions (but not containing their derivatives).
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]
\[ \boxed {y^{\prime }+a \ln \left (x \right ) y^{2}-a f \left (x \right ) \left (x \ln \left (x \right )-x \right ) y=-f \left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a x \ln \left (x \right ) f \left (x \right ) y -a \ln \left (x \right ) y^{2}-f \left (x \right ) a x y -f \left (x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a x \ln \left (x \right ) f \left (x \right ) y -a \ln \left (x \right ) y^{2}-f \left (x \right ) a x y -f \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-f \left (x \right )\), \(f_1(x)=a x \ln \left (x \right ) f \left (x \right )-f \left (x \right ) a x\) and \(f_2(x)=-a \ln \left (x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-a \ln \left (x \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 {a}{x}\\ f_1 f_2 &=-\left (a x \ln \left (x \right ) f \left (x \right )-f \left (x \right ) a x \right ) a \ln \left (x \right )\\ f_2^2 f_0 &=-a^{2} f \left (x \right ) \ln \left (x \right )^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -a \ln \left (x \right ) u^{\prime \prime }\left (x \right )-\left (-\frac {a}{x}-\left (a x \ln \left (x \right ) f \left (x \right )-f \left (x \right ) a x \right ) a \ln \left (x \right )\right ) u^{\prime }\left (x \right )-a^{2} f \left (x \right ) \ln \left (x \right )^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x \left (-1+\ln \left (x \right )\right ) \left (c_{2} \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \right )+c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {x c_{2} \ln \left (x \right ) \left (-1+\ln \left (x \right )\right ) \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \right )+c_{2} {\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}+c_{1} x \ln \left (x \right ) \left (-1+\ln \left (x \right )\right )}{x \left (-1+\ln \left (x \right )\right )} \] Using the above in (1) gives the solution \[ y = \frac {x c_{2} \ln \left (x \right ) \left (-1+\ln \left (x \right )\right ) \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \right )+c_{2} {\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}+c_{1} x \ln \left (x \right ) \left (-1+\ln \left (x \right )\right )}{x^{2} \left (-1+\ln \left (x \right )\right )^{2} a \ln \left (x \right ) \left (c_{2} \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \right )+c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x \ln \left (x \right ) \left (-1+\ln \left (x \right )\right ) \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \right )+c_{3} x \ln \left (x \right )^{2}-c_{3} x \ln \left (x \right )+{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2} a \ln \left (x \right ) \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x +c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x \ln \left (x \right ) \left (-1+\ln \left (x \right )\right ) \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \right )+c_{3} x \ln \left (x \right )^{2}-c_{3} x \ln \left (x \right )+{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2} a \ln \left (x \right ) \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x +c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {x \ln \left (x \right ) \left (-1+\ln \left (x \right )\right ) \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \right )+c_{3} x \ln \left (x \right )^{2}-c_{3} x \ln \left (x \right )+{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2} a \ln \left (x \right ) \left (\int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x +c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+a \ln \left (x \right ) y^{2}-a f \left (x \right ) \left (x \ln \left (x \right )-x \right ) y=-f \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-a \ln \left (x \right ) y^{2}+a f \left (x \right ) \left (x \ln \left (x \right )-x \right ) y-f \left (x \right ) \end {array} \]
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) = (ln(x)^2*f(x)*a*x^2-ln(x)*f(x)*a*x^2+1)*(diff(y(x), x))/(ln(x)*x)-f(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 -> 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)-(-a*ln(x)*y(x)^2+y(x)+(ln(x)*f(x)*a*x-a*x*f(x))*y(x)*x-f(x)*x^2)/x, y(x), explici 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: 0.016 (sec). Leaf size: 227
dsolve(diff(y(x),x)=-a*ln(x)*y(x)^2+a*f(x)*(x*ln(x)-x)*y(x)-f(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-x \left (\ln \left (x \right )-1\right ) {\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}+\left (-2 x^{2} a f \left (x \right )-2\right ) \ln \left (x \right )+x^{2} a f \left (x \right )}{x \left (\ln \left (x \right )-1\right )}d x}+c_{1} a -\left (\int \ln \left (x \right ) {\mathrm e}^{a \left (\int \frac {x f \left (x \right ) \ln \left (x \right )^{2}}{\ln \left (x \right )-1}d x \right )-2 a \left (\int \frac {x f \left (x \right ) \ln \left (x \right )}{\ln \left (x \right )-1}d x \right )+a \left (\int \frac {x f \left (x \right )}{\ln \left (x \right )-1}d x \right )-2 \left (\int \frac {\ln \left (x \right )}{x \left (\ln \left (x \right )-1\right )}d x \right )}d x \right )}{a x \left (\ln \left (x \right )-1\right ) \left (c_{1} a -\left (\int \ln \left (x \right ) {\mathrm e}^{a \left (\int \frac {x f \left (x \right ) \ln \left (x \right )^{2}}{\ln \left (x \right )-1}d x \right )-2 a \left (\int \frac {x f \left (x \right ) \ln \left (x \right )}{\ln \left (x \right )-1}d x \right )+a \left (\int \frac {x f \left (x \right )}{\ln \left (x \right )-1}d x \right )-2 \left (\int \frac {\ln \left (x \right )}{x \left (\ln \left (x \right )-1\right )}d x \right )}d x \right )\right )} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==-a*Log[x]*y[x]^2+a*f[x]*(x*Log[x]-x)*y[x]-f[x],y[x],x,IncludeSingularSolutions -> True]
Not solved