Internal problem ID [10411]
Internal file name [OUTPUT/9359_Monday_June_06_2022_02_18_10_PM_32604344/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: 4.
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]
\[ \boxed {y^{\prime }-\sigma y^{2}-a y=b \,{\mathrm e}^{x}+c} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \sigma \,y^{2}+y a +b \,{\mathrm e}^{x}+c \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \sigma \,y^{2}+y a +b \,{\mathrm e}^{x}+c \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=b \,{\mathrm e}^{x}+c\), \(f_1(x)=a\) and \(f_2(x)=\sigma \). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\sigma 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' &=0\\ f_1 f_2 &=a \sigma \\ f_2^2 f_0 &=\sigma ^{2} \left (b \,{\mathrm e}^{x}+c \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \sigma u^{\prime \prime }\left (x \right )-a \sigma u^{\prime }\left (x \right )+\sigma ^{2} \left (b \,{\mathrm e}^{x}+c \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {x a}{2}} \left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{1} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -c_{1} \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {\left (1+a \right ) x}{2}} \operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )-c_{2} \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {\left (1+a \right ) x}{2}} \operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )+\frac {{\mathrm e}^{\frac {x a}{2}} \left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{1} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{2} \right ) \left (\sqrt {a^{2}-4 c \sigma }+a \right )}{2} \] Using the above in (1) gives the solution \[ y = -\frac {\left (-c_{1} \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {\left (1+a \right ) x}{2}} \operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )-c_{2} \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {\left (1+a \right ) x}{2}} \operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )+\frac {{\mathrm e}^{\frac {x a}{2}} \left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{1} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{2} \right ) \left (\sqrt {a^{2}-4 c \sigma }+a \right )}{2}\right ) {\mathrm e}^{-\frac {x a}{2}}}{\sigma \left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{1} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {2 c_{3} \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}} \operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )+2 \operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}-\left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{3} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )\right ) \left (\sqrt {a^{2}-4 c \sigma }+a \right )}{2 \sigma \left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{3} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 c_{3} \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}} \operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )+2 \operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}-\left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{3} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )\right ) \left (\sqrt {a^{2}-4 c \sigma }+a \right )}{2 \sigma \left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{3} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {2 c_{3} \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}} \operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )+2 \operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }+1, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}-\left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{3} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )\right ) \left (\sqrt {a^{2}-4 c \sigma }+a \right )}{2 \sigma \left (\operatorname {BesselJ}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right ) c_{3} +\operatorname {BesselY}\left (\sqrt {a^{2}-4 c \sigma }, 2 \sqrt {b}\, \sqrt {\sigma }\, {\mathrm e}^{\frac {x}{2}}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\sigma y^{2}-a y=b \,{\mathrm e}^{x}+c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sigma y^{2}+a y+b \,{\mathrm e}^{x}+c \end {array} \]
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) = (diff(y(x), x))*a-sigma*(b*exp(x)+c)*y(x), y(x)` *** Sublevel 2 * 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)] Linear ODE actually solved: (b*sigma*t+c*sigma)*u(t)+(-a*t+t)*diff(u(t),t)+t^2*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: 200
dsolve(diff(y(x),x)=sigma*y(x)^2+a*y(x)+b*exp(x)+c,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {-2 \sqrt {b}\, {\mathrm e}^{\frac {x}{2}} \operatorname {BesselJ}\left (\sqrt {a^{2}-4 \sigma c}+1, 2 \sqrt {\sigma }\, \sqrt {b}\, {\mathrm e}^{\frac {x}{2}}\right ) \sigma -2 \sqrt {b}\, {\mathrm e}^{\frac {x}{2}} \operatorname {BesselY}\left (\sqrt {a^{2}-4 \sigma c}+1, 2 \sqrt {\sigma }\, \sqrt {b}\, {\mathrm e}^{\frac {x}{2}}\right ) c_{1} \sigma +\sqrt {\sigma }\, \left (\operatorname {BesselY}\left (\sqrt {a^{2}-4 \sigma c}, 2 \sqrt {\sigma }\, \sqrt {b}\, {\mathrm e}^{\frac {x}{2}}\right ) c_{1} +\operatorname {BesselJ}\left (\sqrt {a^{2}-4 \sigma c}, 2 \sqrt {\sigma }\, \sqrt {b}\, {\mathrm e}^{\frac {x}{2}}\right )\right ) \left (\sqrt {a^{2}-4 \sigma c}+a \right )}{\sigma ^{\frac {3}{2}} \left (2 \operatorname {BesselY}\left (\sqrt {a^{2}-4 \sigma c}, 2 \sqrt {\sigma }\, \sqrt {b}\, {\mathrm e}^{\frac {x}{2}}\right ) c_{1} +2 \operatorname {BesselJ}\left (\sqrt {a^{2}-4 \sigma c}, 2 \sqrt {\sigma }\, \sqrt {b}\, {\mathrm e}^{\frac {x}{2}}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.971 (sec). Leaf size: 546
DSolve[y'[x]==sigma*y[x]^2+a*y[x]+b*Exp[x]+c,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {a \sqrt {b \sigma e^x} \operatorname {Gamma}\left (\sqrt {a^2-4 c \sigma }+1\right ) \operatorname {BesselJ}\left (\sqrt {a^2-4 c \sigma },2 \sqrt {b e^x \sigma }\right )+b \sigma e^x \operatorname {Gamma}\left (\sqrt {a^2-4 c \sigma }+1\right ) \operatorname {BesselJ}\left (\sqrt {a^2-4 c \sigma }-1,2 \sqrt {b e^x \sigma }\right )-b \sigma e^x \operatorname {Gamma}\left (\sqrt {a^2-4 c \sigma }+1\right ) \operatorname {BesselJ}\left (\sqrt {a^2-4 c \sigma }+1,2 \sqrt {b e^x \sigma }\right )+a c_1 \sqrt {b \sigma e^x} \operatorname {Gamma}\left (1-\sqrt {a^2-4 c \sigma }\right ) \operatorname {BesselJ}\left (-\sqrt {a^2-4 c \sigma },2 \sqrt {b e^x \sigma }\right )+b c_1 \sigma e^x \operatorname {Gamma}\left (1-\sqrt {a^2-4 c \sigma }\right ) \operatorname {BesselJ}\left (-\sqrt {a^2-4 c \sigma }-1,2 \sqrt {b e^x \sigma }\right )-b c_1 \sigma e^x \operatorname {Gamma}\left (1-\sqrt {a^2-4 c \sigma }\right ) \operatorname {BesselJ}\left (1-\sqrt {a^2-4 c \sigma },2 \sqrt {b e^x \sigma }\right )}{2 \sigma \sqrt {b \sigma e^x} \left (\operatorname {Gamma}\left (\sqrt {a^2-4 c \sigma }+1\right ) \operatorname {BesselJ}\left (\sqrt {a^2-4 c \sigma },2 \sqrt {b e^x \sigma }\right )+c_1 \operatorname {Gamma}\left (1-\sqrt {a^2-4 c \sigma }\right ) \operatorname {BesselJ}\left (-\sqrt {a^2-4 c \sigma },2 \sqrt {b e^x \sigma }\right )\right )} \\ y(x)\to \frac {\frac {\sqrt {b \sigma e^x} \left (\operatorname {BesselJ}\left (1-\sqrt {a^2-4 c \sigma },2 \sqrt {b e^x \sigma }\right )-\operatorname {BesselJ}\left (-\sqrt {a^2-4 c \sigma }-1,2 \sqrt {b e^x \sigma }\right )\right )}{\operatorname {BesselJ}\left (-\sqrt {a^2-4 c \sigma },2 \sqrt {b e^x \sigma }\right )}-a}{2 \sigma } \\ \end{align*}