4.18 problem 39

4.18.1 Solving as riccati ode
4.18.2 Maple step by step solution

Internal problem ID [10446]
Internal file name [OUTPUT/9394_Monday_June_06_2022_02_21_28_PM_93629603/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: 39.
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]

\[ \boxed {y^{\prime }-a \,x^{n} y^{2}-y \lambda x=a \,b^{2} x^{n} {\mathrm e}^{\lambda \,x^{2}}} \]

4.18.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,x^{n} y^{2}+\lambda x y +a \,b^{2} x^{n} {\mathrm e}^{\lambda \,x^{2}} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,x^{n} y^{2}+\lambda x y +a \,b^{2} x^{n} {\mathrm e}^{\lambda \,x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \,b^{2} x^{n} {\mathrm e}^{\lambda \,x^{2}}\), \(f_1(x)=\lambda x\) 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 &=x^{n} a \lambda x\\ f_2^2 f_0 &=x^{3 n} a^{3} b^{2} {\mathrm e}^{\lambda \,x^{2}} \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}+x^{n} a \lambda x \right ) u^{\prime }\left (x \right )+x^{3 n} a^{3} b^{2} {\mathrm e}^{\lambda \,x^{2}} u \left (x \right ) &=0 \end {align*}

Solving the above ODE (this ode solved using Maple, not this program), gives

\[ u \left (x \right ) = c_{1} \sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )+c_{2} \cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = a b \,x^{n} {\mathrm e}^{\frac {\lambda \,x^{2}}{2}} \left (c_{1} \cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )-c_{2} \sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )\right ) \] Using the above in (1) gives the solution \[ y = -\frac {b \,{\mathrm e}^{\frac {\lambda \,x^{2}}{2}} \left (c_{1} \cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )-c_{2} \sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )\right )}{c_{1} \sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )+c_{2} \cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = \frac {b \,{\mathrm e}^{\frac {\lambda \,x^{2}}{2}} \left (-c_{3} \cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )+\sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )\right )}{c_{3} \sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )+\cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {b \,{\mathrm e}^{\frac {\lambda \,x^{2}}{2}} \left (-c_{3} \cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )+\sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )\right )}{c_{3} \sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )+\cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )} \\ \end{align*}

Verification of solutions

\[ y = \frac {b \,{\mathrm e}^{\frac {\lambda \,x^{2}}{2}} \left (-c_{3} \cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )+\sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )\right )}{c_{3} \sin \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )+\cos \left (a b \,x^{1+n} 2^{-\frac {1}{2}+\frac {n}{2}} \left (-\lambda \,x^{2}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (\Gamma \left (\frac {1}{2}+\frac {n}{2}\right )-\Gamma \left (\frac {1}{2}+\frac {n}{2}, -\frac {\lambda \,x^{2}}{2}\right )\right )\right )} \] Verified OK.

4.18.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,x^{n} y^{2}-y \lambda x =a \,b^{2} x^{n} {\mathrm e}^{\lambda \,x^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,x^{n} y^{2}+y \lambda x +a \,b^{2} x^{n} {\mathrm e}^{\lambda \,x^{2}} \end {array} \]

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: 90

dsolve(diff(y(x),x)=a*x^n*y(x)^2+lambda*x*y(x)+a*b^2*x^n*exp(lambda*x^2),y(x), singsol=all)
 

\[ y \left (x \right ) = -\tan \left (-a b 2^{\frac {n}{2}-\frac {1}{2}} x^{n +1} \Gamma \left (\frac {n}{2}+\frac {1}{2}\right ) \left (-x^{2} \lambda \right )^{-\frac {n}{2}-\frac {1}{2}}+a b 2^{\frac {n}{2}-\frac {1}{2}} x^{n +1} \left (-x^{2} \lambda \right )^{-\frac {n}{2}-\frac {1}{2}} \Gamma \left (\frac {n}{2}+\frac {1}{2}, -\frac {x^{2} \lambda }{2}\right )+c_{1} \right ) b \,{\mathrm e}^{\frac {x^{2} \lambda }{2}} \]

Solution by Mathematica

Time used: 2.366 (sec). Leaf size: 83

DSolve[y'[x]==a*x^n*y[x]^2+\[Lambda]*x*y[x]+a*b^2*x^n*Exp[\[Lambda]*x^2],y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \sqrt {b^2} e^{\frac {\lambda x^2}{2}} \tan \left (a \sqrt {b^2} \lambda 2^{\frac {n-1}{2}} x^{n+3} \left (\lambda \left (-x^2\right )\right )^{-\frac {n}{2}-\frac {3}{2}} \Gamma \left (\frac {n+1}{2},-\frac {x^2 \lambda }{2}\right )+c_1\right ) \]