6.3 problem 20

Internal problem ID [10467]
Internal file name [OUTPUT/9415_Monday_June_06_2022_02_28_33_PM_45959373/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 1.2.4-2. Equations with hyperbolic tangent and cotangent.
Problem number: 20.
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]

\[ \boxed {y^{\prime }-y^{2}-a x \tanh \left (b x \right )^{m} y=a \tanh \left (b x \right )^{m}} \]

6.3.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+a x \tanh \left (b x \right )^{m} y +a \tanh \left (b x \right )^{m} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+a x \tanh \left (b x \right )^{m} y +a \tanh \left (b x \right )^{m} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \tanh \left (b x \right )^{m}\), \(f_1(x)=\tanh \left (b x \right )^{m} a x\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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' &=0\\ f_1 f_2 &=\tanh \left (b x \right )^{m} a x\\ f_2^2 f_0 &=a \tanh \left (b x \right )^{m} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-\tanh \left (b x \right )^{m} a x u^{\prime }\left (x \right )+a \tanh \left (b x \right )^{m} u \left (x \right ) &=0 \end {align*}

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

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

\[ y = \frac {-c_{3} \left (\int {\mathrm e}^{\int \frac {\tanh \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right )-1-x c_{3} {\mathrm e}^{\int \frac {\tanh \left (b x \right )^{m} a \,x^{2}-2}{x}d x}}{x \left (c_{3} \left (\int {\mathrm e}^{\int \frac {\tanh \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right )+1\right )} \]

Summary

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

Verification of solutions

\[ y = \frac {-c_{3} \left (\int {\mathrm e}^{\int \frac {\tanh \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right )-1-x c_{3} {\mathrm e}^{\int \frac {\tanh \left (b x \right )^{m} a \,x^{2}-2}{x}d x}}{x \left (c_{3} \left (\int {\mathrm e}^{\int \frac {\tanh \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right )+1\right )} \] Verified OK.

6.3.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-a x \tanh \left (b x \right )^{m} y=a \tanh \left (b x \right )^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a x \tanh \left (b x \right )^{m} y+a \tanh \left (b x \right )^{m} \end {array} \]

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 
found: 2 potential symmetries. Proceeding with integration step`
 

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 85

dsolve(diff(y(x),x)=y(x)^2+a*x*tanh(b*x)^m*y(x)+a*tanh(b*x)^m,y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {-{\mathrm e}^{\int \frac {a \tanh \left (b x \right )^{m} x^{2}-2}{x}d x} x -\left (\int {\mathrm e}^{\int \frac {a \tanh \left (b x \right )^{m} x^{2}-2}{x}d x}d x \right )+c_{1}}{\left (-c_{1} +\int {\mathrm e}^{\int \frac {a \tanh \left (b x \right )^{m} x^{2}-2}{x}d x}d x \right ) x} \]

✓ Solution by Mathematica

Time used: 12.331 (sec). Leaf size: 126

DSolve[y'[x]==y[x]^2+a*x*Tanh[b*x]^m*y[x]+a*Tanh[b*x]^m,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -\frac {\exp \left (-\int _1^x-a K[1] \tanh ^m(b K[1])dK[1]\right )+x \int _1^x\frac {\exp \left (-\int _1^{K[2]}-a K[1] \tanh ^m(b K[1])dK[1]\right )}{K[2]^2}dK[2]+c_1 x}{x^2 \left (\int _1^x\frac {\exp \left (-\int _1^{K[2]}-a K[1] \tanh ^m(b K[1])dK[1]\right )}{K[2]^2}dK[2]+c_1\right )} \\ y(x)\to -\frac {1}{x} \\ \end{align*}