Internal problem ID [10441]
Internal file name [OUTPUT/9389_Monday_June_06_2022_02_21_19_PM_91254245/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 1.2.3-2. Equations with power and
exponential functions
Problem number: 34.
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]
\[ \boxed {y^{\prime }-a \,{\mathrm e}^{\lambda x} \left (y-b \,x^{n}-c \right )^{2}=b n \,x^{-1+n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 x^{n} {\mathrm e}^{\lambda x} a b c -2 a b \,x^{n} {\mathrm e}^{\lambda x} y +{\mathrm e}^{\lambda x} a \,c^{2}-2 \,{\mathrm e}^{\lambda x} a c y +{\mathrm e}^{\lambda x} a \,y^{2}+b n \,x^{-1+n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 x^{n} {\mathrm e}^{\lambda x} a b c -2 a b \,x^{n} {\mathrm e}^{\lambda x} y +{\mathrm e}^{\lambda x} a \,c^{2}-2 \,{\mathrm e}^{\lambda x} a c y +{\mathrm e}^{\lambda x} a \,y^{2}+\frac {b n \,x^{n}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 x^{n} {\mathrm e}^{\lambda x} a b c +{\mathrm e}^{\lambda x} a \,c^{2}+b n \,x^{-1+n}\), \(f_1(x)=-2 a b \,x^{n} {\mathrm e}^{\lambda x}-2 \,{\mathrm e}^{\lambda x} a c\) and \(f_2(x)={\mathrm e}^{\lambda x} a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{{\mathrm e}^{\lambda 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' &=a \lambda \,{\mathrm e}^{\lambda x}\\ f_1 f_2 &=\left (-2 a b \,x^{n} {\mathrm e}^{\lambda x}-2 \,{\mathrm e}^{\lambda x} a c \right ) {\mathrm e}^{\lambda x} a\\ f_2^2 f_0 &={\mathrm e}^{2 \lambda x} a^{2} \left (a \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 x^{n} {\mathrm e}^{\lambda x} a b c +{\mathrm e}^{\lambda x} a \,c^{2}+b n \,x^{-1+n}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} {\mathrm e}^{\lambda x} a u^{\prime \prime }\left (x \right )-\left (a \lambda \,{\mathrm e}^{\lambda x}+\left (-2 a b \,x^{n} {\mathrm e}^{\lambda x}-2 \,{\mathrm e}^{\lambda x} a c \right ) {\mathrm e}^{\lambda x} a \right ) u^{\prime }\left (x \right )+{\mathrm e}^{2 \lambda x} a^{2} \left (a \,b^{2} {\mathrm e}^{\lambda x} x^{2 n}+2 x^{n} {\mathrm e}^{\lambda x} a b c +{\mathrm e}^{\lambda x} a \,c^{2}+b n \,x^{-1+n}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\frac {\left (\int \left (2 a \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\lambda \right )d x \right )}{2}} \left (c_{1} \sinh \left (\frac {\lambda x}{2}\right )+c_{2} \cosh \left (\frac {\lambda x}{2}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -{\mathrm e}^{-\frac {\left (\int \left (2 a \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\lambda \right )d x \right )}{2}} \left (\left (a c_{2} \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\frac {\lambda \left (c_{1} +c_{2} \right )}{2}\right ) \cosh \left (\frac {\lambda x}{2}\right )+\sinh \left (\frac {\lambda x}{2}\right ) \left (a c_{1} \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\frac {\lambda \left (c_{1} +c_{2} \right )}{2}\right )\right ) \] Using the above in (1) gives the solution \[ y = \frac {\left (\left (a c_{2} \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\frac {\lambda \left (c_{1} +c_{2} \right )}{2}\right ) \cosh \left (\frac {\lambda x}{2}\right )+\sinh \left (\frac {\lambda x}{2}\right ) \left (a c_{1} \left (b \,x^{n}+c \right ) {\mathrm e}^{\lambda x}-\frac {\lambda \left (c_{1} +c_{2} \right )}{2}\right )\right ) {\mathrm e}^{-\lambda x}}{a \left (c_{1} \sinh \left (\frac {\lambda x}{2}\right )+c_{2} \cosh \left (\frac {\lambda x}{2}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-\lambda \left (c_{3} +1\right ) {\mathrm e}^{-\lambda x}+2 \left (b \,x^{n}+c \right ) a \right ) \cosh \left (\frac {\lambda x}{2}\right )+2 \sinh \left (\frac {\lambda x}{2}\right ) \left (-\frac {\lambda \left (c_{3} +1\right ) {\mathrm e}^{-\lambda x}}{2}+\left (b \,x^{n}+c \right ) c_{3} a \right )}{2 \left (c_{3} \sinh \left (\frac {\lambda x}{2}\right )+\cosh \left (\frac {\lambda x}{2}\right )\right ) a} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-\lambda \left (c_{3} +1\right ) {\mathrm e}^{-\lambda x}+2 \left (b \,x^{n}+c \right ) a \right ) \cosh \left (\frac {\lambda x}{2}\right )+2 \sinh \left (\frac {\lambda x}{2}\right ) \left (-\frac {\lambda \left (c_{3} +1\right ) {\mathrm e}^{-\lambda x}}{2}+\left (b \,x^{n}+c \right ) c_{3} a \right )}{2 \left (c_{3} \sinh \left (\frac {\lambda x}{2}\right )+\cosh \left (\frac {\lambda x}{2}\right )\right ) a} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-\lambda \left (c_{3} +1\right ) {\mathrm e}^{-\lambda x}+2 \left (b \,x^{n}+c \right ) a \right ) \cosh \left (\frac {\lambda x}{2}\right )+2 \sinh \left (\frac {\lambda x}{2}\right ) \left (-\frac {\lambda \left (c_{3} +1\right ) {\mathrm e}^{-\lambda x}}{2}+\left (b \,x^{n}+c \right ) c_{3} a \right )}{2 \left (c_{3} \sinh \left (\frac {\lambda x}{2}\right )+\cosh \left (\frac {\lambda x}{2}\right )\right ) a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,{\mathrm e}^{\lambda x} \left (y-b \,x^{n}-c \right )^{2}=b n \,x^{-1+n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,{\mathrm e}^{\lambda x} \left (y-b \,x^{n}-c \right )^{2}+b n \,x^{-1+n} \end {array} \]
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: 50
dsolve(diff(y(x),x)=a*exp(lambda*x)*(y(x)-b*x^n-c)^2+b*n*x^(n-1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {a c_{1} \lambda \left (b \,x^{n}+c \right ) {\mathrm e}^{x \lambda }+x^{n} a b -c_{1} \lambda ^{2}+a c}{\left (\lambda c_{1} {\mathrm e}^{x \lambda }+1\right ) a} \]
✓ Solution by Mathematica
Time used: 1.563 (sec). Leaf size: 40
DSolve[y'[x]==a*Exp[\[Lambda]*x]*(y[x]-b*x^n-c)^2+b*n*x^(n-1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\lambda }{-a e^{\lambda x}+c_1 \lambda }+b x^n+c \\ y(x)\to b x^n+c \\ \end{align*}