Internal problem ID [10436]
Internal file name [OUTPUT/9384_Monday_June_06_2022_02_21_04_PM_3481108/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: 29.
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]
\[ \boxed {y^{\prime }-a \,x^{n} y^{2}-\lambda y=-a \,b^{2} x^{n} {\mathrm e}^{2 \lambda x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,x^{n} y^{2}+y \lambda -a \,b^{2} x^{n} {\mathrm e}^{2 \lambda x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,x^{n} y^{2}+y \lambda -a \,b^{2} x^{n} {\mathrm e}^{2 \lambda x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a \,b^{2} x^{n} {\mathrm e}^{2 \lambda x}\), \(f_1(x)=\lambda \) and \(f_2(x)=x^{n} a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{x^{n} 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' &=\frac {x^{n} n a}{x}\\ f_1 f_2 &=\lambda \,x^{n} a\\ f_2^2 f_0 &=-x^{3 n} a^{3} b^{2} {\mathrm e}^{2 \lambda x} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} x^{n} a u^{\prime \prime }\left (x \right )-\left (\frac {x^{n} n a}{x}+\lambda \,x^{n} a \right ) u^{\prime }\left (x \right )-x^{3 n} a^{3} b^{2} {\mathrm e}^{2 \lambda x} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = -c_{1} \sinh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )+c_{2} \cosh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = a b \,x^{n} {\mathrm e}^{\lambda x} \left (c_{2} \sinh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )-c_{1} \cosh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )\right ) \] Using the above in (1) gives the solution \[ y = -\frac {b \,{\mathrm e}^{\lambda x} \left (c_{2} \sinh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )-c_{1} \cosh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )\right )}{-c_{1} \sinh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )+c_{2} \cosh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {b \,{\mathrm e}^{\lambda x} \left (c_{3} \cosh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )-\sinh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )\right )}{-c_{3} \sinh \left (\frac {\left (-\lambda x \right )^{-n} x^{n} a b \left ({\mathrm e}^{\lambda x} \left (-\lambda x \right )^{n}+\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right )}{\lambda }\right )+\cosh \left (\frac {\left (-\lambda x \right )^{-n} x^{n} a b \left ({\mathrm e}^{\lambda x} \left (-\lambda x \right )^{n}+\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right )}{\lambda }\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {b \,{\mathrm e}^{\lambda x} \left (c_{3} \cosh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )-\sinh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )\right )}{-c_{3} \sinh \left (\frac {\left (-\lambda x \right )^{-n} x^{n} a b \left ({\mathrm e}^{\lambda x} \left (-\lambda x \right )^{n}+\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right )}{\lambda }\right )+\cosh \left (\frac {\left (-\lambda x \right )^{-n} x^{n} a b \left ({\mathrm e}^{\lambda x} \left (-\lambda x \right )^{n}+\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right )}{\lambda }\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {b \,{\mathrm e}^{\lambda x} \left (c_{3} \cosh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )-\sinh \left (\frac {b \,x^{n} a \left (\left (\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right ) \left (-\lambda x \right )^{-n}+{\mathrm e}^{\lambda x}\right )}{\lambda }\right )\right )}{-c_{3} \sinh \left (\frac {\left (-\lambda x \right )^{-n} x^{n} a b \left ({\mathrm e}^{\lambda x} \left (-\lambda x \right )^{n}+\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right )}{\lambda }\right )+\cosh \left (\frac {\left (-\lambda x \right )^{-n} x^{n} a b \left ({\mathrm e}^{\lambda x} \left (-\lambda x \right )^{n}+\Gamma \left (n , -\lambda x \right ) n -\Gamma \left (1+n \right )\right )}{\lambda }\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,x^{n} y^{2}-\lambda y=-a \,b^{2} x^{n} {\mathrm e}^{2 \lambda x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,x^{n} y^{2}+\lambda y-a \,b^{2} x^{n} {\mathrm e}^{2 \lambda x} \end {array} \]
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 <- Chini successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 62
dsolve(diff(y(x),x)=a*x^n*y(x)^2+lambda*y(x)-a*b^2*x^n*exp(2*lambda*x),y(x), singsol=all)
\[ y \left (x \right ) = \tanh \left (\frac {-a b \,x^{n} \left (n \Gamma \left (n , -x \lambda \right )-\Gamma \left (n +1\right )\right ) \left (-x \lambda \right )^{-n}-b a \,{\mathrm e}^{x \lambda } x^{n}+i \lambda c_{1}}{\lambda }\right ) b \,{\mathrm e}^{x \lambda } \]
✓ Solution by Mathematica
Time used: 1.69 (sec). Leaf size: 57
DSolve[y'[x]==a*x^n*y[x]^2+\[Lambda]*y[x]-a*b^2*x^n*Exp[2*\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {-b^2} e^{\lambda x} \tan \left (\frac {a \sqrt {-b^2} x^n (\lambda (-x))^{-n} \Gamma (n+1,-x \lambda )}{\lambda }+c_1\right ) \]