Internal problem ID [10426]
Internal file name [OUTPUT/9374_Monday_June_06_2022_02_19_43_PM_62152854/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: 19.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime }-{\mathrm e}^{\mu x} \left (y-b \,{\mathrm e}^{\lambda x}\right )^{2}=b \lambda \,{\mathrm e}^{\lambda x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= {\mathrm e}^{\mu x} {\mathrm e}^{2 \lambda x} b^{2}-2 \,{\mathrm e}^{\lambda x} {\mathrm e}^{\mu x} b y +{\mathrm e}^{\mu x} y^{2}+b \lambda \,{\mathrm e}^{\lambda x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = {\mathrm e}^{\mu x} {\mathrm e}^{2 \lambda x} b^{2}-2 \,{\mathrm e}^{\lambda x} {\mathrm e}^{\mu x} b y +{\mathrm e}^{\mu x} y^{2}+b \lambda \,{\mathrm e}^{\lambda x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)={\mathrm e}^{\mu x} {\mathrm e}^{2 \lambda x} b^{2}+b \lambda \,{\mathrm e}^{\lambda x}\), \(f_1(x)=-2 b \,{\mathrm e}^{\lambda x} {\mathrm e}^{\mu x}\) and \(f_2(x)={\mathrm e}^{\mu x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{{\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' &=\mu \,{\mathrm e}^{\mu x}\\ f_1 f_2 &=-2 b \,{\mathrm e}^{\lambda x} {\mathrm e}^{2 \mu x}\\ f_2^2 f_0 &={\mathrm e}^{2 \mu x} \left ({\mathrm e}^{\mu x} {\mathrm e}^{2 \lambda x} b^{2}+b \lambda \,{\mathrm e}^{\lambda x}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} {\mathrm e}^{\mu x} u^{\prime \prime }\left (x \right )-\left (\mu \,{\mathrm e}^{\mu x}-2 b \,{\mathrm e}^{\lambda x} {\mathrm e}^{2 \mu x}\right ) u^{\prime }\left (x \right )+{\mathrm e}^{2 \mu x} \left ({\mathrm e}^{\mu x} {\mathrm e}^{2 \lambda x} b^{2}+b \lambda \,{\mathrm e}^{\lambda x}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {-2 \,{\mathrm e}^{x \left (\lambda +\mu \right )} b +\mu x \left (\lambda +\mu \right )}{2 \lambda +2 \mu }} \left (c_{1} \sinh \left (\frac {\mu x}{2}\right )+c_{2} \cosh \left (\frac {\mu x}{2}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -{\mathrm e}^{\frac {-2 \,{\mathrm e}^{x \left (\lambda +\mu \right )} b +\mu x \left (\lambda +\mu \right )}{2 \lambda +2 \mu }} \left (\left ({\mathrm e}^{x \left (\lambda +\mu \right )} c_{2} b -\frac {\mu \left (c_{1} +c_{2} \right )}{2}\right ) \cosh \left (\frac {\mu x}{2}\right )+\sinh \left (\frac {\mu x}{2}\right ) \left ({\mathrm e}^{x \left (\lambda +\mu \right )} b c_{1} -\frac {\mu \left (c_{1} +c_{2} \right )}{2}\right )\right ) \] Using the above in (1) gives the solution \[ y = \frac {\left (\left ({\mathrm e}^{x \left (\lambda +\mu \right )} c_{2} b -\frac {\mu \left (c_{1} +c_{2} \right )}{2}\right ) \cosh \left (\frac {\mu x}{2}\right )+\sinh \left (\frac {\mu x}{2}\right ) \left ({\mathrm e}^{x \left (\lambda +\mu \right )} b c_{1} -\frac {\mu \left (c_{1} +c_{2} \right )}{2}\right )\right ) {\mathrm e}^{-\mu x}}{c_{1} \sinh \left (\frac {\mu x}{2}\right )+c_{2} \cosh \left (\frac {\mu x}{2}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\left ({\mathrm e}^{x \left (\lambda +\mu \right )} b -\frac {\mu \left (c_{3} +1\right )}{2}\right ) \cosh \left (\frac {\mu x}{2}\right )+\sinh \left (\frac {\mu x}{2}\right ) \left ({\mathrm e}^{x \left (\lambda +\mu \right )} b c_{3} -\frac {\mu \left (c_{3} +1\right )}{2}\right )\right ) {\mathrm e}^{-\mu x}}{c_{3} \sinh \left (\frac {\mu x}{2}\right )+\cosh \left (\frac {\mu x}{2}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\left ({\mathrm e}^{x \left (\lambda +\mu \right )} b -\frac {\mu \left (c_{3} +1\right )}{2}\right ) \cosh \left (\frac {\mu x}{2}\right )+\sinh \left (\frac {\mu x}{2}\right ) \left ({\mathrm e}^{x \left (\lambda +\mu \right )} b c_{3} -\frac {\mu \left (c_{3} +1\right )}{2}\right )\right ) {\mathrm e}^{-\mu x}}{c_{3} \sinh \left (\frac {\mu x}{2}\right )+\cosh \left (\frac {\mu x}{2}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\left ({\mathrm e}^{x \left (\lambda +\mu \right )} b -\frac {\mu \left (c_{3} +1\right )}{2}\right ) \cosh \left (\frac {\mu x}{2}\right )+\sinh \left (\frac {\mu x}{2}\right ) \left ({\mathrm e}^{x \left (\lambda +\mu \right )} b c_{3} -\frac {\mu \left (c_{3} +1\right )}{2}\right )\right ) {\mathrm e}^{-\mu x}}{c_{3} \sinh \left (\frac {\mu x}{2}\right )+\cosh \left (\frac {\mu x}{2}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-{\mathrm e}^{\mu x} \left (y-b \,{\mathrm e}^{\lambda x}\right )^{2}=b \lambda \,{\mathrm e}^{\lambda x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{\mu x} \left (y-b \,{\mathrm e}^{\lambda x}\right )^{2}+b \lambda \,{\mathrm e}^{\lambda x} \end {array} \]
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: <- Riccati particular polynomial solution successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 44
dsolve(diff(y(x),x)=exp(mu*x)*(y(x)-b*exp(lambda*x))^2+b*lambda*exp(lambda*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left ({\mathrm e}^{x \left (\lambda +\mu \right )} c_{1} b \mu +b \,{\mathrm e}^{x \lambda }-c_{1} \mu ^{2}\right ) {\mathrm e}^{-x \mu }}{c_{1} \mu +{\mathrm e}^{-x \mu }} \]
✓ Solution by Mathematica
Time used: 1.524 (sec). Leaf size: 40
DSolve[y'[x]==Exp[\[Mu]*x]*(y[x]-b*Exp[\[Lambda]*x])^2+b*\[Lambda]*Exp[\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to b e^{\lambda x}+\frac {\mu }{-e^{\mu x}+c_1 \mu } \\ y(x)\to b e^{\lambda x} \\ \end{align*}