Internal problem ID [10432]
Internal file name [OUTPUT/9380_Monday_June_06_2022_02_20_55_PM_76225887/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: 25.
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]
\[ \boxed {y^{\prime }-{\mathrm e}^{\lambda x} y^{2}-a \,x^{n} y=a \lambda \,x^{n} {\mathrm e}^{-\lambda x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= {\mathrm e}^{\lambda x} y^{2}+x^{n} a y +a \lambda \,x^{n} {\mathrm e}^{-\lambda x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = {\mathrm e}^{\lambda x} y^{2}+x^{n} a y +a \lambda \,x^{n} {\mathrm e}^{-\lambda x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \lambda \,x^{n} {\mathrm e}^{-\lambda x}\), \(f_1(x)=x^{n} a\) and \(f_2(x)={\mathrm e}^{\lambda x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{{\mathrm e}^{\lambda x} 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' &=\lambda \,{\mathrm e}^{\lambda x}\\ f_1 f_2 &={\mathrm e}^{\lambda x} x^{n} a\\ f_2^2 f_0 &={\mathrm e}^{2 \lambda x} a \lambda \,x^{n} {\mathrm e}^{-\lambda x} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} {\mathrm e}^{\lambda x} u^{\prime \prime }\left (x \right )-\left ({\mathrm e}^{\lambda x} x^{n} a +\lambda \,{\mathrm e}^{\lambda x}\right ) u^{\prime }\left (x \right )+{\mathrm e}^{2 \lambda x} a \lambda \,x^{n} {\mathrm e}^{-\lambda x} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\int \frac {\left (\int {\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}d x \right ) \lambda -c_{1} \lambda +{\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}}{\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x -c_{1}}d x} c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {c_{2} \left (\left (\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x \right ) \lambda -c_{1} \lambda +{\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}\right ) {\mathrm e}^{\int \frac {\left (\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x \right ) \lambda -c_{1} \lambda +{\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}}{\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x -c_{1}}d x}}{\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x -c_{1}} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\left (\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x \right ) \lambda -c_{1} \lambda +{\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}\right ) {\mathrm e}^{\int \frac {\left (\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x \right ) \lambda -c_{1} \lambda +{\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}}{\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x -c_{1}}d x} {\mathrm e}^{-\lambda x} {\mathrm e}^{\int -\frac {\int \lambda \,{\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}d x +{\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}-c_{1} \lambda }{\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x -c_{1}}d x}}{\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x -c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {{\mathrm e}^{-\lambda x} \left (\left (\int {\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}d x \right ) \lambda -\lambda c_{3} +{\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}\right )}{\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x -c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {{\mathrm e}^{-\lambda x} \left (\left (\int {\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}d x \right ) \lambda -\lambda c_{3} +{\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}\right )}{\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x -c_{3}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {{\mathrm e}^{-\lambda x} \left (\left (\int {\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}d x \right ) \lambda -\lambda c_{3} +{\mathrm e}^{\frac {a \,x^{1+n}-\lambda x \left (1+n \right )}{1+n}}\right )}{\int {\mathrm e}^{\frac {x \left (x^{n} a -\lambda \left (1+n \right )\right )}{1+n}}d x -c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-{\mathrm e}^{\lambda x} y^{2}-a \,x^{n} y=a \lambda \,x^{n} {\mathrm e}^{-\lambda x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{\lambda x} y^{2}+a \,x^{n} y+a \lambda \,x^{n} {\mathrm e}^{-\lambda x} \end {array} \]
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) = (x^n*a+lambda)*(diff(y(x), x))-exp(lambda*x)*a*lambda*x^n*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 -> 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)-(exp(lambda*x)*y(x)^2+y(x)+a*x^n*y(x)*x+x^2*a*lambda*x^n*exp(-lambda*x))/x, y(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`[0, exp(-1/(n+1)*a*x^(n+1)+2*x*lambda)*(y+lambda/exp(x*lambda))^2]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 89
dsolve(diff(y(x),x)=exp(lambda*x)*y(x)^2+a*x^(n)*y(x)+a*lambda*x^n*exp(-lambda*x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {{\mathrm e}^{-x \lambda } \left (\left (\int {\mathrm e}^{\frac {x^{n +1} a -x \lambda \left (n +1\right )}{n +1}}d x \right ) \lambda +\lambda c_{1} +{\mathrm e}^{\frac {x^{n +1} a -x \lambda \left (n +1\right )}{n +1}}\right )}{c_{1} +\int {\mathrm e}^{\frac {x \left (a \,x^{n}-\left (n +1\right ) \lambda \right )}{n +1}}d x} \]
✓ Solution by Mathematica
Time used: 1.93 (sec). Leaf size: 254
DSolve[y'[x]==Exp[\[Lambda]*x]*y[x]^2+a*x^(n)*y[x]+a*\[Lambda]*x^n*Exp[-\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (\frac {e^{\frac {a x^{n+1}}{n+1}}}{\left (\lambda +e^{x \lambda } K[2]\right )^2}-\int _1^x\left (\frac {2 e^{\frac {a K[1]^{n+1}}{n+1}} \left (a \lambda K[1]^n+a e^{\lambda K[1]} K[2] K[1]^n+e^{2 \lambda K[1]} K[2]^2\right )}{\left (\lambda +e^{\lambda K[1]} K[2]\right )^3}-\frac {e^{\frac {a K[1]^{n+1}}{n+1}-\lambda K[1]} \left (a e^{\lambda K[1]} K[1]^n+2 e^{2 \lambda K[1]} K[2]\right )}{\left (\lambda +e^{\lambda K[1]} K[2]\right )^2}\right )dK[1]\right )dK[2]+\int _1^x-\frac {e^{\frac {a K[1]^{n+1}}{n+1}-\lambda K[1]} \left (a \lambda K[1]^n+a e^{\lambda K[1]} y(x) K[1]^n+e^{2 \lambda K[1]} y(x)^2\right )}{\left (\lambda +e^{\lambda K[1]} y(x)\right )^2}dK[1]=c_1,y(x)\right ] \]