Internal problem ID [10416]
Internal file name [OUTPUT/9364_Monday_June_06_2022_02_18_59_PM_79365723/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: 9.
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]
\[ \boxed {y^{\prime }-a \,{\mathrm e}^{k x} y^{2}=b \,{\mathrm e}^{s x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,{\mathrm e}^{k x} y^{2}+b \,{\mathrm e}^{s x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,{\mathrm e}^{k x} y^{2}+b \,{\mathrm e}^{s x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=b \,{\mathrm e}^{s x}\), \(f_1(x)=0\) 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 &=0\\ f_2^2 f_0 &={\mathrm e}^{2 k x} {\mathrm e}^{s x} a^{2} b \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \,{\mathrm e}^{k x} u^{\prime \prime }\left (x \right )-a k \,{\mathrm e}^{k x} u^{\prime }\left (x \right )+{\mathrm e}^{2 k x} {\mathrm e}^{s x} a^{2} b u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = -\frac {{\mathrm e}^{-\frac {s x}{2}} \left (\sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}} \operatorname {BesselJ}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{1} +\operatorname {BesselY}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}} c_{2} -s \left (c_{1} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )+c_{2} \operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )\right )}{\sqrt {a}\, \sqrt {b}} \] The above shows that \[ u^{\prime }\left (x \right ) = \left (-c_{1} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )-c_{2} \operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right ) \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (2 k +s \right )}{2}} \] Using the above in (1) gives the solution \[ y = \frac {\left (-c_{1} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )-c_{2} \operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right ) b \,{\mathrm e}^{\frac {x \left (2 k +s \right )}{2}} {\mathrm e}^{-k x} {\mathrm e}^{\frac {s x}{2}}}{\sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}} \operatorname {BesselJ}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{1} +\operatorname {BesselY}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}} c_{2} -s \left (c_{1} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )+c_{2} \operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {b \,{\mathrm e}^{s x} \left (c_{3} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )+\operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )}{\sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}} \operatorname {BesselJ}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}-s \left (c_{3} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )+\operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {b \,{\mathrm e}^{s x} \left (c_{3} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )+\operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )}{\sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}} \operatorname {BesselJ}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}-s \left (c_{3} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )+\operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {b \,{\mathrm e}^{s x} \left (c_{3} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )+\operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )}{\sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}} \operatorname {BesselJ}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {k +2 s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}-s \left (c_{3} \operatorname {BesselJ}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )+\operatorname {BesselY}\left (\frac {s}{k +s}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,{\mathrm e}^{k x} y^{2}=b \,{\mathrm e}^{s x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,{\mathrm e}^{k x} y^{2}+b \,{\mathrm e}^{s x} \end {array} \]
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) = k*(diff(y(x), x))-a*exp(k*x)*b*exp(s*x)*y(x), y(x)` *** Sublevel 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 quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful Change of variables used: [x = ln(t)/(s+k)] Linear ODE actually solved: a*b*u(t)+(k*s+s^2)*diff(u(t),t)+(k^2*t+2*k*s*t+s^2*t)*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 228
dsolve(diff(y(x),x)=a*exp(k*x)*y(x)^2+b*exp(s*x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {b \,{\mathrm e}^{s x} \left (\operatorname {BesselY}\left (\frac {s}{s +k}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {s}{s +k}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right )\right )}{\operatorname {BesselJ}\left (\frac {2 s +k}{s +k}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right ) \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}+\sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}} \operatorname {BesselY}\left (\frac {2 s +k}{s +k}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right ) c_{1} -s \left (\operatorname {BesselY}\left (\frac {s}{s +k}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {s}{s +k}, \frac {2 \sqrt {a}\, \sqrt {b}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right )\right )} \]
✓ Solution by Mathematica
Time used: 6.491 (sec). Leaf size: 1097
DSolve[y'[x]==a*Exp[k*x]*y[x]^2+b*Exp[s*x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {e^{-k x} \left (-k K_{\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}}\left (2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )-c_1 k (-1)^{\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}} \operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )}{(k+s)^2},2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+(k+s) \sqrt {-\frac {a b \log ^2\left (e^{k+s}\right ) \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}}}{(k+s)^4}} \left (K_{\frac {k \log \left (e^{k+s}\right )-(k+s)^2}{(k+s)^2}}\left (2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+K_{\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}+1}\left (2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )-c_1 (-1)^{\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}} \left (\operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )-(k+s)^2}{(k+s)^2},2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+\operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}+1,2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )\right )\right )\right )}{2 a \left (K_{\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}}\left (2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+c_1 (-1)^{\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}} \operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )}{(k+s)^2},2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )\right )} \\ y(x)\to \frac {e^{-k x} \left (-\frac {(k+s) \sqrt {-\frac {a b \log ^2\left (e^{k+s}\right ) \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}}}{(k+s)^4}} \left (\operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}-1,2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+\operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}+1,2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )\right )}{\operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )}{(k+s)^2},2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )}-k\right )}{2 a} \\ y(x)\to \frac {e^{-k x} \left (-\frac {(k+s) \sqrt {-\frac {a b \log ^2\left (e^{k+s}\right ) \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}}}{(k+s)^4}} \left (\operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}-1,2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+\operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )}{(k+s)^2}+1,2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )\right )}{\operatorname {BesselI}\left (\frac {k \log \left (e^{k+s}\right )}{(k+s)^2},2 \sqrt {-\frac {a b \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )}-k\right )}{2 a} \\ \end{align*}