Internal problem ID [10608]
Internal file name [OUTPUT/9556_Monday_June_06_2022_03_08_46_PM_10138442/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: 16.
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]
\[ \boxed {y^{\prime }-f \left (x \right ) y^{2}+f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right ) y=a \lambda \,{\mathrm e}^{\lambda x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -a \,{\mathrm e}^{\lambda x} f \left (x \right ) y -f \left (x \right ) b y +f \left (x \right ) y^{2}+a \lambda \,{\mathrm e}^{\lambda x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -a \,{\mathrm e}^{\lambda x} f \left (x \right ) y -f \left (x \right ) b y +f \left (x \right ) y^{2}+a \lambda \,{\mathrm e}^{\lambda x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \lambda \,{\mathrm e}^{\lambda x}\), \(f_1(x)=-a \,{\mathrm e}^{\lambda x} f \left (x \right )-f \left (x \right ) b\) and \(f_2(x)=f \left (x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{f \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' &=f^{\prime }\left (x \right )\\ f_1 f_2 &=\left (-a \,{\mathrm e}^{\lambda x} f \left (x \right )-f \left (x \right ) b \right ) f \left (x \right )\\ f_2^2 f_0 &=\lambda \,{\mathrm e}^{\lambda x} f \left (x \right )^{2} a \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} f \left (x \right ) u^{\prime \prime }\left (x \right )-\left (f^{\prime }\left (x \right )+\left (-a \,{\mathrm e}^{\lambda x} f \left (x \right )-f \left (x \right ) b \right ) f \left (x \right )\right ) u^{\prime }\left (x \right )+\lambda \,{\mathrm e}^{\lambda x} f \left (x \right )^{2} a u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\left (\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x \right )} \left (c_{1} +\left (\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x \right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -f \left (x \right ) \left (a \left (c_{1} +\left (\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x \right ) c_{2} \right ) {\mathrm e}^{\lambda x -\left (\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x \right )}+{\mathrm e}^{-\left (\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x \right )} \left (\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x \right ) c_{2} b +{\mathrm e}^{-\left (\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x \right )} c_{1} b -c_{2} \right ) \] Using the above in (1) gives the solution \[ y = \frac {\left (a \left (c_{1} +\left (\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x \right ) c_{2} \right ) {\mathrm e}^{\lambda x -\left (\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x \right )}+{\mathrm e}^{-\left (\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x \right )} \left (\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x \right ) c_{2} b +{\mathrm e}^{-\left (\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x \right )} c_{1} b -c_{2} \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}}{c_{1} +\left (\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x \right ) c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (a \,{\mathrm e}^{\lambda x}+b \right ) \left (\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x \right )+a c_{3} {\mathrm e}^{\lambda x}+b c_{3} -{\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}}{c_{3} +\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (a \,{\mathrm e}^{\lambda x}+b \right ) \left (\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x \right )+a c_{3} {\mathrm e}^{\lambda x}+b c_{3} -{\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}}{c_{3} +\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (a \,{\mathrm e}^{\lambda x}+b \right ) \left (\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x \right )+a c_{3} {\mathrm e}^{\lambda x}+b c_{3} -{\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}}{c_{3} +\int f \left (x \right ) {\mathrm e}^{\int f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right )d x}d x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-f \left (x \right ) y^{2}+f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right ) y=a \lambda \,{\mathrm e}^{\lambda x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=f \left (x \right ) y^{2}-f \left (x \right ) \left (a \,{\mathrm e}^{\lambda x}+b \right ) y+a \lambda \,{\mathrm e}^{\lambda x} \end {array} \]
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) = (-exp(lambda*x)*f(x)^2*a-f(x)^2*b+diff(f(x), x))*(diff(y(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)-(f(x)*y(x)^2+y(x)+(-a*exp(lambda*x)*f(x)-f(x)*b)*y(x)*x+x^2*a*lambda*exp(lambda*x 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`
✗ Solution by Maple
dsolve(diff(y(x),x)=f(x)*y(x)^2-f(x)*(a*exp(lambda*x)+b)*y(x)+a*lambda*exp(lambda*x),y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==f[x]*y[x]^2-f[x]*(a*Exp[\[Lambda]*x]+b)*y[x]+a*\[Lambda]*Exp[\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
Not solved