Internal problem ID [10419]
Internal file name [OUTPUT/9367_Monday_June_06_2022_02_19_04_PM_92605250/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: 12.
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]
\[ \boxed {y^{\prime }-a \,{\mathrm e}^{\mu x} y^{2}-\lambda y=-a \,b^{2} {\mathrm e}^{\left (\mu +2 \lambda \right ) x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,{\mathrm e}^{\mu x} y^{2}+y \lambda -a \,b^{2} {\mathrm e}^{\left (\mu +2 \lambda \right ) x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,{\mathrm e}^{\mu x} y^{2}+y \lambda -a \,b^{2} {\mathrm e}^{2 \lambda x} {\mathrm e}^{\mu x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a \,b^{2} {\mathrm e}^{\left (\mu +2 \lambda \right ) x}\), \(f_1(x)=\lambda \) and \(f_2(x)=a \,{\mathrm e}^{\mu x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a \,{\mathrm e}^{\mu 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' &=a \mu \,{\mathrm e}^{\mu x}\\ f_1 f_2 &=\lambda a \,{\mathrm e}^{\mu x}\\ f_2^2 f_0 &=-a^{3} {\mathrm e}^{2 \mu x} b^{2} {\mathrm e}^{\left (\mu +2 \lambda \right ) x} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \,{\mathrm e}^{\mu x} u^{\prime \prime }\left (x \right )-\left (a \mu \,{\mathrm e}^{\mu x}+\lambda a \,{\mathrm e}^{\mu x}\right ) u^{\prime }\left (x \right )-a^{3} {\mathrm e}^{2 \mu x} b^{2} {\mathrm e}^{\left (\mu +2 \lambda \right ) x} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )+c_{2} \cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {a b \,{\mathrm e}^{2 x \left (\lambda +\mu \right )} \left (-c_{1} \cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )+c_{2} \sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )\right )}{\sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}} \] Using the above in (1) gives the solution \[ y = -\frac {b \,{\mathrm e}^{2 x \left (\lambda +\mu \right )} \left (-c_{1} \cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )+c_{2} \sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )\right ) {\mathrm e}^{-\mu x}}{\sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}\, \left (c_{1} \sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )+c_{2} \cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {b \,{\mathrm e}^{\left (\mu +2 \lambda \right ) x} \left (c_{3} \cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )-\sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )\right )}{\sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}\, \left (c_{3} \sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )+\cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {b \,{\mathrm e}^{\left (\mu +2 \lambda \right ) x} \left (c_{3} \cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )-\sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )\right )}{\sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}\, \left (c_{3} \sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )+\cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {b \,{\mathrm e}^{\left (\mu +2 \lambda \right ) x} \left (c_{3} \cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )-\sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )\right )}{\sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}\, \left (c_{3} \sin \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )+\cos \left (\frac {a b \sqrt {-{\mathrm e}^{2 \lambda x} {\mathrm e}^{2 \mu x}}}{\lambda +\mu }\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,{\mathrm e}^{\mu x} y^{2}-\lambda y=-a \,b^{2} {\mathrm e}^{\left (\mu +2 \lambda \right ) x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,{\mathrm e}^{\mu x} y^{2}+\lambda y-a \,b^{2} {\mathrm e}^{\left (\mu +2 \lambda \right ) x} \end {array} \]
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 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) = (lambda+mu)*(diff(y(x), x))+a^2*exp(x*mu)*b^2*exp(2*lambda*x+mu*x)*y(x Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] <- linear_1 successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 79
dsolve(diff(y(x),x)=a*exp(mu*x)*y(x)^2+lambda*y(x)-a*b^2*exp((mu+2*lambda)*x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {b \left (c_{1} \sinh \left (\frac {a b \,{\mathrm e}^{x \left (\lambda +\mu \right )}}{\lambda +\mu }\right )+\cosh \left (\frac {a b \,{\mathrm e}^{x \left (\lambda +\mu \right )}}{\lambda +\mu }\right )\right ) {\mathrm e}^{x \lambda }}{c_{1} \cosh \left (\frac {a b \,{\mathrm e}^{x \left (\lambda +\mu \right )}}{\lambda +\mu }\right )+\sinh \left (\frac {a b \,{\mathrm e}^{x \left (\lambda +\mu \right )}}{\lambda +\mu }\right )} \]
✓ Solution by Mathematica
Time used: 2.706 (sec). Leaf size: 286
DSolve[y'[x]==a*Exp[\[Mu]*x]*y[x]^2+\[Lambda]*y[x]-a*b^2*Exp[(\[Mu]+2*\[Lambda])*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \fbox {$-\frac {\tan \left (\frac {a b^2 e^{x (2 \lambda +\mu )} \sqrt {-\frac {e^{-2 x \lambda }}{b^2}}}{\lambda +\mu }-c_1\right )}{\sqrt {-\frac {e^{-2 x \lambda }}{b^2}}}\text { if }\text {condition}$} \]