Internal problem ID [10425]
Internal file name [OUTPUT/9373_Monday_June_06_2022_02_19_41_PM_60059753/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. Equations Containing
Exponential Functions
Problem number: 18.
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]
\[ \boxed {y^{\prime }-a \,{\mathrm e}^{k x} y^{2}-y b=c \,{\mathrm e}^{k n x}+d \,{\mathrm e}^{k \left (1+2 n \right ) x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,{\mathrm e}^{k x} y^{2}+y b +c \,{\mathrm e}^{k n x}+d \,{\mathrm e}^{k \left (1+2 n \right ) x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,{\mathrm e}^{k x} y^{2}+y b +c \,{\mathrm e}^{k n x}+d \,{\mathrm e}^{2 k n x} {\mathrm e}^{k x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c \,{\mathrm e}^{k n x}+d \,{\mathrm e}^{k \left (1+2 n \right ) x}\), \(f_1(x)=b\) and \(f_2(x)=a \,{\mathrm e}^{k x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a \,{\mathrm e}^{k 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' &=a k \,{\mathrm e}^{k x}\\ f_1 f_2 &=b a \,{\mathrm e}^{k x}\\ f_2^2 f_0 &=a^{2} {\mathrm e}^{2 k x} \left (c \,{\mathrm e}^{k n x}+d \,{\mathrm e}^{k \left (1+2 n \right ) x}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \,{\mathrm e}^{k x} u^{\prime \prime }\left (x \right )-\left (b a \,{\mathrm e}^{k x}+a k \,{\mathrm e}^{k x}\right ) u^{\prime }\left (x \right )+a^{2} {\mathrm e}^{2 k x} \left (c \,{\mathrm e}^{k n x}+d \,{\mathrm e}^{k \left (1+2 n \right ) x}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) {\mathrm e}^{-k x}}{a \operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) {\mathrm e}^{-k x}}{a \operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) {\mathrm e}^{-k x}}{a \operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) {\mathrm e}^{-k x}}{a \operatorname {DESol}\left (\left \{{\mathrm e}^{2 k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a d +{\mathrm e}^{k x \left (1+n \right )} \textit {\_Y} \left (x \right ) a c +\textit {\_Y}^{\prime \prime }\left (x \right )+\left (-b -k \right ) \textit {\_Y}^{\prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,{\mathrm e}^{k x} y^{2}-y b =c \,{\mathrm e}^{k n x}+d \,{\mathrm e}^{k \left (1+2 n \right ) x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,{\mathrm e}^{k x} y^{2}+y b +c \,{\mathrm e}^{k n x}+d \,{\mathrm e}^{k \left (1+2 n \right ) x} \end {array} \]
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) = (b+k)*(diff(y(x), x))-a*exp(k*x)*(c*exp(k*n*x)+d*exp(2*k*n*x+k*x))*y(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 <- 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 <- 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*exp(k*x)*y(x)^2+y(x)+y(x)*b*x+x^2*(c*exp(k*n*x)+d*exp(2*k*n*x+k*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`
✗ Solution by Maple
dsolve(diff(y(x),x)=a*exp(k*x)*y(x)^2+b*y(x)+c*exp(k*n*x)+d*exp(k*(2*n+1)*x),y(x), singsol=all)
\[ \text {No solution found} \]
✓ Solution by Mathematica
Time used: 27.598 (sec). Leaf size: 2503
DSolve[y'[x]==a*Exp[k*x]*y[x]^2+b*y[x]+c*Exp[k*n*x]+d*Exp[k*(2*n+1)*x],y[x],x,IncludeSingularSolutions -> True]
Too large to display