Internal problem ID [10438]
Internal file name [OUTPUT/9386_Monday_June_06_2022_02_21_08_PM_74422371/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 1.2.3-2. Equations with power and
exponential functions
Problem number: 31.
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]
\[ \boxed {y^{\prime }+\left (k +1\right ) x^{k} y^{2}-a \,x^{k +1} {\mathrm e}^{\lambda x} y=-{\mathrm e}^{\lambda x} a} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -x^{k} y^{2} k +a \,x^{k +1} {\mathrm e}^{\lambda x} y -x^{k} y^{2}-{\mathrm e}^{\lambda x} a \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -x^{k} y^{2} k +a \,x^{k} x \,{\mathrm e}^{\lambda x} y -x^{k} y^{2}-{\mathrm e}^{\lambda x} a \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-{\mathrm e}^{\lambda x} a\), \(f_1(x)=a \,x^{k +1} {\mathrm e}^{\lambda x}\) and \(f_2(x)=-x^{k} k -x^{k}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\left (-x^{k} k -x^{k}\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 {k^{2} x^{k}}{x}-\frac {k \,x^{k}}{x}\\ f_1 f_2 &=a \,x^{k +1} {\mathrm e}^{\lambda x} \left (-x^{k} k -x^{k}\right )\\ f_2^2 f_0 &=-\left (-x^{k} k -x^{k}\right )^{2} {\mathrm e}^{\lambda x} a \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \left (-x^{k} k -x^{k}\right ) u^{\prime \prime }\left (x \right )-\left (-\frac {k^{2} x^{k}}{x}-\frac {k \,x^{k}}{x}+a \,x^{k +1} {\mathrm e}^{\lambda x} \left (-x^{k} k -x^{k}\right )\right ) u^{\prime }\left (x \right )-\left (-x^{k} k -x^{k}\right )^{2} {\mathrm e}^{\lambda x} a u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{k +1} \left (\left (\int x^{-2 k -2} {\mathrm e}^{\int \left (a \,x^{k +1} {\mathrm e}^{\lambda x}+\frac {k}{x}\right )d x}d x \right ) c_{2} +c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = x^{k} \left (k +1\right ) \left (\left (\int {\mathrm e}^{\int \frac {a \,x^{2+k} {\mathrm e}^{\lambda x}+k}{x}d x} x^{-2 k -2}d x \right ) c_{2} +c_{1} \right )+c_{2} x^{-k -1} {\mathrm e}^{\int \frac {a \,x^{2+k} {\mathrm e}^{\lambda x}+k}{x}d x} \] Using the above in (1) gives the solution \[ y = -\frac {\left (x^{k} \left (k +1\right ) \left (\left (\int {\mathrm e}^{\int \frac {a \,x^{2+k} {\mathrm e}^{\lambda x}+k}{x}d x} x^{-2 k -2}d x \right ) c_{2} +c_{1} \right )+c_{2} x^{-k -1} {\mathrm e}^{\int \frac {a \,x^{2+k} {\mathrm e}^{\lambda x}+k}{x}d x}\right ) x^{-k -1}}{\left (-x^{k} k -x^{k}\right ) \left (\left (\int x^{-2 k -2} {\mathrm e}^{\int \left (a \,x^{k +1} {\mathrm e}^{\lambda x}+\frac {k}{x}\right )d x}d x \right ) c_{2} +c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\int x^{-2 k -2} {\mathrm e}^{\int \left (a \,x^{k +1} {\mathrm e}^{\lambda x}+\frac {k}{x}\right )d x}d x +c_{3} \right ) \left (k +1\right ) x^{-k -1}+x^{-3 k -2} {\mathrm e}^{\int \left (a \,x^{k +1} {\mathrm e}^{\lambda x}+\frac {k}{x}\right )d x}}{\left (k +1\right ) \left (\int {\mathrm e}^{\int \frac {a \,x^{2+k} {\mathrm e}^{\lambda x}+k}{x}d x} x^{-2 k -2}d x +c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\int x^{-2 k -2} {\mathrm e}^{\int \left (a \,x^{k +1} {\mathrm e}^{\lambda x}+\frac {k}{x}\right )d x}d x +c_{3} \right ) \left (k +1\right ) x^{-k -1}+x^{-3 k -2} {\mathrm e}^{\int \left (a \,x^{k +1} {\mathrm e}^{\lambda x}+\frac {k}{x}\right )d x}}{\left (k +1\right ) \left (\int {\mathrm e}^{\int \frac {a \,x^{2+k} {\mathrm e}^{\lambda x}+k}{x}d x} x^{-2 k -2}d x +c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\int x^{-2 k -2} {\mathrm e}^{\int \left (a \,x^{k +1} {\mathrm e}^{\lambda x}+\frac {k}{x}\right )d x}d x +c_{3} \right ) \left (k +1\right ) x^{-k -1}+x^{-3 k -2} {\mathrm e}^{\int \left (a \,x^{k +1} {\mathrm e}^{\lambda x}+\frac {k}{x}\right )d x}}{\left (k +1\right ) \left (\int {\mathrm e}^{\int \frac {a \,x^{2+k} {\mathrm e}^{\lambda x}+k}{x}d x} x^{-2 k -2}d x +c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\left (k +1\right ) x^{k} y^{2}-a \,x^{k +1} {\mathrm e}^{\lambda x} y=-{\mathrm e}^{\lambda x} a \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\left (k +1\right ) x^{k} y^{2}+a \,x^{k +1} {\mathrm e}^{\lambda x} y-{\mathrm e}^{\lambda x} a \end {array} \]
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^(1+k)*exp(lambda*x)*a*x+k)*(diff(y(x), x))/x-x^k*(1+k)*a*exp(lambda 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^k*k-x^k)*y(x)^2+y(x)+a*x^(1+k)*exp(lambda*x)*y(x)*x-x^2*exp(lambda*x)*a)/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
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 184
dsolve(diff(y(x),x)=-(k+1)*x^k*y(x)^2+a*x^(k+1)*exp(lambda*x)*y(x)-a*exp(lambda*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{-1-k} \left (x^{1+k} {\mathrm e}^{\int \frac {x^{1+k} {\mathrm e}^{x \lambda } a x -2 k -2}{x}d x}+\left (\int x^{k} {\mathrm e}^{a \left (\int x^{1+k} {\mathrm e}^{x \lambda }d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (1+k \right )}d x \right ) k +\int x^{k} {\mathrm e}^{a \left (\int x^{1+k} {\mathrm e}^{x \lambda }d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (1+k \right )}d x -c_{1} \right )}{\left (\int x^{k} {\mathrm e}^{a \left (\int x^{1+k} {\mathrm e}^{x \lambda }d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (1+k \right )}d x \right ) k +\int x^{k} {\mathrm e}^{a \left (\int x^{1+k} {\mathrm e}^{x \lambda }d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (1+k \right )}d x -c_{1}} \]
✓ Solution by Mathematica
Time used: 86.249 (sec). Leaf size: 280
DSolve[y'[x]==-(k+1)*x^k*y[x]^2+a*x^(k+1)*Exp[\[Lambda]*x]*y[x]-a*Exp[\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {a \lambda \exp \left (\frac {a \left (-\log \left (e^{\lambda x}\right )\right )^{-k} \left (\frac {\log \left (e^{\lambda x}\right )}{\lambda }\right )^k \Gamma \left (k+2,-\log \left (e^{x \lambda }\right )\right )}{\lambda ^2}\right ) \left (1+c_1 \int _1^{e^{x \lambda }}\exp \left (-\frac {a \Gamma (k+2,-\log (K[1])) (-\log (K[1]))^{-k} \left (\frac {\log (K[1])}{\lambda }\right )^k}{\lambda ^2}\right )dK[1]\right )}{a c_1 \lambda \left (\frac {\log \left (e^{\lambda x}\right )}{\lambda }\right )^{k+1} \exp \left (\frac {a \left (-\log \left (e^{\lambda x}\right )\right )^{-k} \left (\frac {\log \left (e^{\lambda x}\right )}{\lambda }\right )^k \Gamma \left (k+2,-\log \left (e^{x \lambda }\right )\right )}{\lambda ^2}\right ) \int _1^{e^{x \lambda }}\exp \left (-\frac {a \Gamma (k+2,-\log (K[1])) (-\log (K[1]))^{-k} \left (\frac {\log (K[1])}{\lambda }\right )^k}{\lambda ^2}\right )dK[1]+a \lambda \left (\frac {\log \left (e^{\lambda x}\right )}{\lambda }\right )^{k+1} \exp \left (\frac {a \left (-\log \left (e^{\lambda x}\right )\right )^{-k} \left (\frac {\log \left (e^{\lambda x}\right )}{\lambda }\right )^k \Gamma \left (k+2,-\log \left (e^{x \lambda }\right )\right )}{\lambda ^2}\right )-c_1 \lambda ^2} \]