Internal problem ID [10409]
Internal file name [OUTPUT/9357_Monday_June_06_2022_02_18_05_PM_73417862/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: 2.
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]
\[ \boxed {y^{\prime }-y^{2}=a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x}\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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 &=0\\ f_2^2 f_0 &=a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\left (a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} \left (c_{1} +\operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\left ({\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }} \lambda c_{2} +{\mathrm e}^{\lambda x} \operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{2} a +{\mathrm e}^{\lambda x} c_{1} a \right ) {\mathrm e}^{-\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }} \] Using the above in (1) gives the solution \[ y = \frac {{\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }} \lambda c_{2} +{\mathrm e}^{\lambda x} \operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{2} a +{\mathrm e}^{\lambda x} c_{1} a}{c_{1} +\operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {{\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }} \lambda +\operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) {\mathrm e}^{\lambda x} a +{\mathrm e}^{\lambda x} c_{3} a}{c_{3} +\operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }} \lambda +\operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) {\mathrm e}^{\lambda x} a +{\mathrm e}^{\lambda x} c_{3} a}{c_{3} +\operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }} \lambda +\operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) {\mathrm e}^{\lambda x} a +{\mathrm e}^{\lambda x} c_{3} a}{c_{3} +\operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{\lambda x}}{\lambda }\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}=a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a \lambda \,{\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{2 \lambda x} \end {array} \]
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 Special trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (-a*lambda*exp(lambda*x)+a^2*exp(2*lambda*x))*y(x), y(x)` *** Sub 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 <- Kovacics algorithm successful Change of variables used: [x = ln(t)/lambda] Linear ODE actually solved: (-a^2*t+a*lambda)*u(t)+lambda^2*diff(u(t),t)+lambda^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: 63
dsolve(diff(y(x),x)=y(x)^2+a*lambda*exp(lambda*x)-a^2*exp(2*lambda*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{x \lambda } \operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{x \lambda }}{\lambda }\right ) c_{1} a +{\mathrm e}^{\frac {2 a \,{\mathrm e}^{x \lambda }}{\lambda }} c_{1} \lambda +{\mathrm e}^{x \lambda } a}{\operatorname {expIntegral}_{1}\left (-\frac {2 a \,{\mathrm e}^{x \lambda }}{\lambda }\right ) c_{1} +1} \]
✓ Solution by Mathematica
Time used: 2.507 (sec). Leaf size: 79
DSolve[y'[x]==y[x]^2+a*\[Lambda]*Exp[\[Lambda]*x]-a^2*Exp[2*\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {a e^{\lambda x} \operatorname {ExpIntegralEi}\left (\frac {2 a e^{x \lambda }}{\lambda }\right )+\lambda \left (-e^{\frac {2 a e^{\lambda x}}{\lambda }}\right )+a c_1 e^{\lambda x}}{\operatorname {ExpIntegralEi}\left (\frac {2 a e^{x \lambda }}{\lambda }\right )+c_1} \\ y(x)\to a e^{\lambda x} \\ \end{align*}