problem 10

Internal problem ID [10457]
Internal ﬁle name [OUTPUT/9405_Monday_June_06_2022_02_24_32_PM_77442611/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 1.2.4-1. Equations with hyperbolic sine and cosine
Problem number: 10.
ODE order: 1.
ODE degree: 1.

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

5.10.1 Solving as riccati ode

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

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

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

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

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

Summary

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

Veriﬁcation of solutions

5.10.2 Maple step by step solution

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

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`

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

\begin{align*} y(x)\to -\frac {\int _1^x\frac {\exp \left (-\frac {2^{-m} a \left (e^{-b K[1]}+e^{b K[1]}\right )^m \left (1+e^{2 b K[1]}\right )^{-m} \left (\, _3F_2\left (-m,-\frac {m}{2},-\frac {m}{2};1-\frac {m}{2},1-\frac {m}{2};-e^{2 b K[1]}\right )+b m \operatorname {Hypergeometric2F1}\left (-m,-\frac {m}{2},1-\frac {m}{2},-e^{2 b K[1]}\right ) K[1]\right )}{b^2 m^2}\right )}{K[1]^2}dK[1]+\frac {\exp \left (-\frac {a 2^{-m} \left (e^{-b x}+e^{b x}\right )^m \left (e^{2 b x}+1\right )^{-m} \left (\, _3F_2\left (-m,-\frac {m}{2},-\frac {m}{2};1-\frac {m}{2},1-\frac {m}{2};-e^{2 b x}\right )+b m x \operatorname {Hypergeometric2F1}\left (-m,-\frac {m}{2},1-\frac {m}{2},-e^{2 b x}\right )\right )}{b^2 m^2}\right )}{x}+c_1}{x \left (\int _1^x\frac {\exp \left (-\frac {2^{-m} a \left (e^{-b K[1]}+e^{b K[1]}\right )^m \left (1+e^{2 b K[1]}\right )^{-m} \left (\, _3F_2\left (-m,-\frac {m}{2},-\frac {m}{2};1-\frac {m}{2},1-\frac {m}{2};-e^{2 b K[1]}\right )+b m \operatorname {Hypergeometric2F1}\left (-m,-\frac {m}{2},1-\frac {m}{2},-e^{2 b K[1]}\right ) K[1]\right )}{b^2 m^2}\right )}{K[1]^2}dK[1]+c_1\right )} \\ y(x)\to -\frac {1}{x} \\ \end{align*}

5.10 problem 10

5.10.1 Solving as riccati ode

5.10.2 Maple step by step solution