Internal problem ID [10424]
Internal file name [OUTPUT/9372_Monday_June_06_2022_02_19_40_PM_7352493/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: 17.
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]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-a \,{\mathrm e}^{\left (\mu +2 \lambda \right ) x} y^{2}-\left ({\mathrm e}^{x \left (\lambda +\mu \right )} b -\lambda \right ) y=c \,{\mathrm e}^{\mu x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,{\mathrm e}^{\left (\mu +2 \lambda \right ) x} y^{2}+{\mathrm e}^{x \left (\lambda +\mu \right )} b y +c \,{\mathrm e}^{\mu x}-y \lambda \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,{\mathrm e}^{2 \lambda x} {\mathrm e}^{\mu x} y^{2}+{\mathrm e}^{\lambda x} {\mathrm e}^{\mu x} b y +c \,{\mathrm e}^{\mu x}-y \lambda \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c \,{\mathrm e}^{\mu x}\), \(f_1(x)={\mathrm e}^{x \left (\lambda +\mu \right )} b -\lambda \) and \(f_2(x)={\mathrm e}^{\left (\mu +2 \lambda \right ) x} a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{{\mathrm e}^{\left (\mu +2 \lambda \right ) x} a 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' &=\left (\mu +2 \lambda \right ) {\mathrm e}^{\left (\mu +2 \lambda \right ) x} a\\ f_1 f_2 &=\left ({\mathrm e}^{x \left (\lambda +\mu \right )} b -\lambda \right ) {\mathrm e}^{\left (\mu +2 \lambda \right ) x} a\\ f_2^2 f_0 &={\mathrm e}^{2 \left (\mu +2 \lambda \right ) x} a^{2} c \,{\mathrm e}^{\mu x} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} {\mathrm e}^{\left (\mu +2 \lambda \right ) x} a u^{\prime \prime }\left (x \right )-\left (\left (\mu +2 \lambda \right ) {\mathrm e}^{\left (\mu +2 \lambda \right ) x} a +\left ({\mathrm e}^{x \left (\lambda +\mu \right )} b -\lambda \right ) {\mathrm e}^{\left (\mu +2 \lambda \right ) x} a \right ) u^{\prime }\left (x \right )+{\mathrm e}^{2 \left (\mu +2 \lambda \right ) x} a^{2} c \,{\mathrm e}^{\mu x} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives Unable to solve. Terminating.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,{\mathrm e}^{\left (\mu +2 \lambda \right ) x} y^{2}-\left ({\mathrm e}^{x \left (\lambda +\mu \right )} b -\lambda \right ) y=c \,{\mathrm e}^{\mu x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,{\mathrm e}^{\left (\mu +2 \lambda \right ) x} y^{2}+\left ({\mathrm e}^{x \left (\lambda +\mu \right )} b -\lambda \right ) y+c \,{\mathrm e}^{\mu x} \end {array} \]
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 <- Chini successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 79
dsolve(diff(y(x),x)=a*exp((2*lambda+mu)*x)*y(x)^2+(b*exp((lambda+mu)*x)-lambda)*y(x)+c*exp(mu*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-x \lambda } \left (\sqrt {4 a \,b^{2} c -b^{4}}\, \tan \left (\frac {\left ({\mathrm e}^{x \left (\lambda +\mu \right )} b +\left (\lambda +\mu \right ) c_{1} \right ) \sqrt {4 a \,b^{2} c -b^{4}}}{2 b^{2} \left (\lambda +\mu \right )}\right )-b^{2}\right )}{2 a b} \]
✓ Solution by Mathematica
Time used: 6.375 (sec). Leaf size: 349
DSolve[y'[x]==a*Exp[(2*\[Lambda]+\[Mu])*x]*y[x]^2+(b*Exp[(\[Lambda]+\[Mu])*x]-\[Lambda])*y[x]+c*Exp[\[Mu]*x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {e^{\lambda (-x)} \left (b^2 e^{x (\lambda +\mu )} \left (\pi +i c_1 \left (e^{\sqrt {\frac {\left (b^2-4 a c\right ) e^{2 x (\lambda +\mu )}}{(\lambda +\mu )^2}}}-1\right )\right )-b (\lambda +\mu ) \sqrt {\frac {\left (b^2-4 a c\right ) e^{2 x (\lambda +\mu )}}{(\lambda +\mu )^2}} \left (\pi -i c_1 \left (e^{\sqrt {\frac {\left (b^2-4 a c\right ) e^{2 x (\lambda +\mu )}}{(\lambda +\mu )^2}}}+1\right )\right )-4 a c e^{x (\lambda +\mu )} \left (\pi +i c_1 \left (e^{\sqrt {\frac {\left (b^2-4 a c\right ) e^{2 x (\lambda +\mu )}}{(\lambda +\mu )^2}}}-1\right )\right )\right )}{2 a (\lambda +\mu ) \sqrt {\frac {\left (b^2-4 a c\right ) e^{2 x (\lambda +\mu )}}{(\lambda +\mu )^2}} \left (\pi -i c_1 \left (e^{\sqrt {\frac {\left (b^2-4 a c\right ) e^{2 x (\lambda +\mu )}}{(\lambda +\mu )^2}}}+1\right )\right )} \\ y(x)\to \frac {e^{\lambda (-x)} \left (-(\lambda +\mu ) e^{-x (\lambda +\mu )} \sqrt {\frac {\left (b^2-4 a c\right ) e^{2 x (\lambda +\mu )}}{(\lambda +\mu )^2}} \tanh \left (\frac {1}{2} \sqrt {\frac {\left (b^2-4 a c\right ) e^{2 x (\lambda +\mu )}}{(\lambda +\mu )^2}}\right )-b\right )}{2 a} \\ \end{align*}