Internal problem ID [10413]
Internal file name [OUTPUT/9361_Monday_June_06_2022_02_18_12_PM_24173088/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: 6.
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]
\[ \boxed {y^{\prime }-y^{2}-a \,{\mathrm e}^{\lambda x} y=-a \,{\mathrm e}^{\lambda x} b -b^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+a \,{\mathrm e}^{\lambda x} y -a \,{\mathrm e}^{\lambda x} b -b^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+a \,{\mathrm e}^{\lambda x} y -a \,{\mathrm e}^{\lambda x} b -b^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a \,{\mathrm e}^{\lambda x} b -b^{2}\), \(f_1(x)={\mathrm e}^{\lambda x} a\) 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 &={\mathrm e}^{\lambda x} a\\ f_2^2 f_0 &=-a \,{\mathrm e}^{\lambda x} b -b^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-{\mathrm e}^{\lambda x} a u^{\prime }\left (x \right )+\left (-a \,{\mathrm e}^{\lambda x} b -b^{2}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {-\lambda ^{2} x +{\mathrm e}^{\lambda x} a}{2 \lambda }} \left (\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{1} +\operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = 2 \left (c_{2} \left ({\mathrm e}^{\lambda x} a +\frac {3 b}{2}-\lambda \right ) \operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )-\frac {c_{1} b \left (-2 \,{\mathrm e}^{\frac {a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \left (\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )^{\frac {2 b +\lambda }{2 \lambda }}+\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )\right )}{2}\right ) {\mathrm e}^{\frac {-\lambda ^{2} x +{\mathrm e}^{\lambda x} a}{2 \lambda }} \] Using the above in (1) gives the solution \[ y = -\frac {2 \left (c_{2} \left ({\mathrm e}^{\lambda x} a +\frac {3 b}{2}-\lambda \right ) \operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )-\frac {c_{1} b \left (-2 \,{\mathrm e}^{\frac {a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \left (\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )^{\frac {2 b +\lambda }{2 \lambda }}+\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )\right )}{2}\right )}{\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{1} +\operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-2 \,{\mathrm e}^{\lambda x} a -3 b +2 \lambda \right ) \operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )-2 \left ({\mathrm e}^{\frac {a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \left (\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )^{\frac {2 b +\lambda }{2 \lambda }}-\frac {\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )}{2}\right ) b c_{3}}{\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{3} +\operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-2 \,{\mathrm e}^{\lambda x} a -3 b +2 \lambda \right ) \operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )-2 \left ({\mathrm e}^{\frac {a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \left (\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )^{\frac {2 b +\lambda }{2 \lambda }}-\frac {\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )}{2}\right ) b c_{3}}{\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{3} +\operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-2 \,{\mathrm e}^{\lambda x} a -3 b +2 \lambda \right ) \operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )-2 \left ({\mathrm e}^{\frac {a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \left (\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )^{\frac {2 b +\lambda }{2 \lambda }}-\frac {\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )}{2}\right ) b c_{3}}{\operatorname {WhittakerM}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right ) c_{3} +\operatorname {WhittakerW}\left (\frac {-2 b +\lambda }{2 \lambda }, \frac {b}{\lambda }, \frac {a \,{\mathrm e}^{\lambda x}}{\lambda }\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-a \,{\mathrm e}^{\lambda x} y=-a \,{\mathrm e}^{\lambda x} b -b^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a \,{\mathrm e}^{\lambda x} y-a \,{\mathrm e}^{\lambda x} b -b^{2} \end {array} \]
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: <- Riccati particular case Kamke (b) successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 73
dsolve(diff(y(x),x)=y(x)^2+a*exp(lambda*x)*y(x)-a*b*exp(lambda*x)-b^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-b \left (\int {\mathrm e}^{\frac {2 b \lambda x +{\mathrm e}^{x \lambda } a}{\lambda }}d x \right )+c_{1} b +{\mathrm e}^{\frac {2 b \lambda x +{\mathrm e}^{x \lambda } a}{\lambda }}}{-\left (\int {\mathrm e}^{\frac {2 b \lambda x +{\mathrm e}^{x \lambda } a}{\lambda }}d x \right )+c_{1}} \]
✓ Solution by Mathematica
Time used: 0.944 (sec). Leaf size: 115
DSolve[y'[x]==y[x]^2+a*Exp[\[Lambda]*x]*y[x]-a*b*Exp[\[Lambda]*x]-b^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {b \left (-2 \lambda e^{\frac {a e^{\lambda x}}{\lambda }} \left (-\frac {a e^{\lambda x}}{\lambda }\right )^{\frac {2 b}{\lambda }}+2 b \Gamma \left (\frac {2 b}{\lambda },0,-\frac {a e^{x \lambda }}{\lambda }\right )+c_1 \lambda (-1)^{b/\lambda }\right )}{2 b \Gamma \left (\frac {2 b}{\lambda },0,-\frac {a e^{x \lambda }}{\lambda }\right )+c_1 \lambda (-1)^{b/\lambda }} \\ y(x)\to b \\ \end{align*}