Internal problem ID [10420]
Internal file name [OUTPUT/9368_Monday_June_06_2022_02_19_05_PM_49258887/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: 13.
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]
\[ \boxed {y^{\prime }-{\mathrm e}^{\lambda x} y^{2}-a \,{\mathrm e}^{\mu x} y=a \lambda \,{\mathrm e}^{\left (\mu -\lambda \right ) x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= {\mathrm e}^{\lambda x} y^{2}+a \,{\mathrm e}^{\mu x} y +a \lambda \,{\mathrm e}^{\left (\mu -\lambda \right ) x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = {\mathrm e}^{\lambda x} y^{2}+a \,{\mathrm e}^{\mu x} y +a \lambda \,{\mathrm e}^{-\lambda x} {\mathrm e}^{\mu x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \lambda \,{\mathrm e}^{\left (\mu -\lambda \right ) x}\), \(f_1(x)=a \,{\mathrm e}^{\mu x}\) 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 &=a \,{\mathrm e}^{\mu x} {\mathrm e}^{\lambda x}\\ f_2^2 f_0 &={\mathrm e}^{2 \lambda x} a \lambda \,{\mathrm e}^{\left (\mu -\lambda \right ) 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 (a \,{\mathrm e}^{\mu x} {\mathrm e}^{\lambda x}+\lambda \,{\mathrm e}^{\lambda x}\right ) u^{\prime }\left (x \right )+{\mathrm e}^{2 \lambda x} a \lambda \,{\mathrm e}^{\left (\mu -\lambda \right ) x} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{\lambda x}+c_{2} \operatorname {hypergeom}\left (\left [-\frac {\lambda }{\mu }\right ], \left [\frac {\mu -\lambda }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\left (c_{2} \operatorname {hypergeom}\left (\left [\frac {\mu -\lambda }{\mu }\right ], \left [\frac {-\lambda +2 \mu }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right ) a \,{\mathrm e}^{\mu x}-{\mathrm e}^{\lambda x} c_{1} \left (\mu -\lambda \right )\right ) \lambda }{\mu -\lambda } \] Using the above in (1) gives the solution \[ y = \frac {\left (c_{2} \operatorname {hypergeom}\left (\left [\frac {\mu -\lambda }{\mu }\right ], \left [\frac {-\lambda +2 \mu }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right ) a \,{\mathrm e}^{\mu x}-{\mathrm e}^{\lambda x} c_{1} \left (\mu -\lambda \right )\right ) \lambda \,{\mathrm e}^{-\lambda x}}{\left (\mu -\lambda \right ) \left (c_{1} {\mathrm e}^{\lambda x}+c_{2} \operatorname {hypergeom}\left (\left [-\frac {\lambda }{\mu }\right ], \left [\frac {\mu -\lambda }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\lambda \left (\operatorname {hypergeom}\left (\left [\frac {\mu -\lambda }{\mu }\right ], \left [\frac {-\lambda +2 \mu }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right ) a \,{\mathrm e}^{\left (\mu -\lambda \right ) x}-c_{3} \left (\mu -\lambda \right )\right )}{\left (\mu -\lambda \right ) \left (c_{3} {\mathrm e}^{\lambda x}+\operatorname {hypergeom}\left (\left [-\frac {\lambda }{\mu }\right ], \left [\frac {\mu -\lambda }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\lambda \left (\operatorname {hypergeom}\left (\left [\frac {\mu -\lambda }{\mu }\right ], \left [\frac {-\lambda +2 \mu }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right ) a \,{\mathrm e}^{\left (\mu -\lambda \right ) x}-c_{3} \left (\mu -\lambda \right )\right )}{\left (\mu -\lambda \right ) \left (c_{3} {\mathrm e}^{\lambda x}+\operatorname {hypergeom}\left (\left [-\frac {\lambda }{\mu }\right ], \left [\frac {\mu -\lambda }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\lambda \left (\operatorname {hypergeom}\left (\left [\frac {\mu -\lambda }{\mu }\right ], \left [\frac {-\lambda +2 \mu }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right ) a \,{\mathrm e}^{\left (\mu -\lambda \right ) x}-c_{3} \left (\mu -\lambda \right )\right )}{\left (\mu -\lambda \right ) \left (c_{3} {\mathrm e}^{\lambda x}+\operatorname {hypergeom}\left (\left [-\frac {\lambda }{\mu }\right ], \left [\frac {\mu -\lambda }{\mu }\right ], \frac {a \,{\mathrm e}^{\mu x}}{\mu }\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-{\mathrm e}^{\lambda x} y^{2}-a \,{\mathrm e}^{\mu x} y=a \lambda \,{\mathrm e}^{\left (\mu -\lambda \right ) x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{\lambda x} y^{2}+a \,{\mathrm e}^{\mu x} y+a \lambda \,{\mathrm e}^{\left (\mu -\lambda \right ) x} \end {array} \]
Maple trace Kovacic algorithm successful
`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) = (a*exp(x*mu)+lambda)*(diff(y(x), x))-exp(lambda*x)*a*lambda*exp(-lambd 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 A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre <- Kummer successful <- special function solution successful Solution using Kummer functions still has integrals. Trying a hypergeometric solution. -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE -> Trying to convert hypergeometric functions to elementary form... <- elementary form could result into a too large expression - returning special function form of solution, free of un <- Kovacics algorithm successful Change of variables used: [x = ln(t)/mu] Linear ODE actually solved: a*lambda*u(t)+(-a*mu*t-lambda*mu+mu^2)*diff(u(t),t)+mu^2*t*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 97
dsolve(diff(y(x),x)=exp(lambda*x)*y(x)^2+a*exp(mu*x)*y(x)+a*lambda*exp((mu-lambda)*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\lambda \left (a c_{1} {\mathrm e}^{\left (\mu -\lambda \right ) x} \operatorname {hypergeom}\left (\left [\frac {\mu -\lambda }{\mu }\right ], \left [\frac {-\lambda +2 \mu }{\mu }\right ], \frac {a \,{\mathrm e}^{x \mu }}{\mu }\right )+\lambda -\mu \right )}{\left (\mu -\lambda \right ) \left (c_{1} \operatorname {hypergeom}\left (\left [-\frac {\lambda }{\mu }\right ], \left [\frac {\mu -\lambda }{\mu }\right ], \frac {a \,{\mathrm e}^{x \mu }}{\mu }\right )+{\mathrm e}^{x \lambda }\right )} \]
✓ Solution by Mathematica
Time used: 4.392 (sec). Leaf size: 148
DSolve[y'[x]==Exp[\[Lambda]*x]*y[x]^2+a*Exp[\[Mu]*x]*y[x]+a*\[Lambda]*Exp[(\[Mu]-\[Lambda])*x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {e^{\lambda (-x)} \left (-\lambda \left (-\frac {a e^{\mu x}}{\mu }\right )^{\lambda /\mu } \Gamma \left (-\frac {\lambda }{\mu },-\frac {a e^{x \mu }}{\mu }\right )+\mu e^{\frac {a e^{\mu x}}{\mu }}+c_1 \lambda \left (e^{\mu x}\right )^{\lambda /\mu }\right )}{-\left (-\frac {a e^{\mu x}}{\mu }\right )^{\lambda /\mu } \Gamma \left (-\frac {\lambda }{\mu },-\frac {a e^{x \mu }}{\mu }\right )+c_1 \left (e^{\mu x}\right )^{\lambda /\mu }} \\ y(x)\to \lambda \left (-e^{\lambda (-x)}\right ) \\ \end{align*}