problem 1

Internal problem ID [10330]
Internal ﬁle name [OUTPUT/9278_Monday_June_06_2022_01_45_23_PM_48991554/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. 1.2.2. Equations Containing Power Functions
Problem number: 1.
ODE order: 1.
ODE degree: 1.

2.1.1 Solving as riccati ode

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

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

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

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

\[ y = \frac {\left (\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b x +c \right )}{b}\right ) c_{3} +\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b x +c \right )}{b}\right )\right ) \left (a b \right )^{\frac {1}{3}}}{a \left (c_{3} \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b x +c \right )}{b}\right )+\operatorname {AiryBi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b x +c \right )}{b}\right )\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b x +c \right )}{b}\right ) c_{3} +\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b x +c \right )}{b}\right )\right ) \left (a b \right )^{\frac {1}{3}}}{a \left (c_{3} \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b x +c \right )}{b}\right )+\operatorname {AiryBi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b x +c \right )}{b}\right )\right )} \\ \end{align*}

Veriﬁcation of solutions

2.1.2 Maple step by step solution

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

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 
differential order: 1; looking for linear symmetries 
trying exact 
Looking for potential symmetries 
trying Riccati 
trying Riccati Special 
trying Riccati sub-methods: 
   <- Abel AIR successful: ODE belongs to the 0F1 0-parameter (Airy type) class`

\[ y \left (x \right ) = \frac {\left (\frac {b}{\sqrt {a}}\right )^{\frac {1}{3}} \left (\operatorname {AiryAi}\left (1, -\frac {b x +c}{\left (\frac {b}{\sqrt {a}}\right )^{\frac {2}{3}}}\right ) c_{1} +\operatorname {AiryBi}\left (1, -\frac {b x +c}{\left (\frac {b}{\sqrt {a}}\right )^{\frac {2}{3}}}\right )\right )}{\sqrt {a}\, \left (c_{1} \operatorname {AiryAi}\left (-\frac {b x +c}{\left (\frac {b}{\sqrt {a}}\right )^{\frac {2}{3}}}\right )+\operatorname {AiryBi}\left (-\frac {b x +c}{\left (\frac {b}{\sqrt {a}}\right )^{\frac {2}{3}}}\right )\right )} \]

\begin{align*} y(x)\to \frac {b \left (\operatorname {AiryBiPrime}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )+c_1 \operatorname {AiryAiPrime}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )\right )}{(-a b)^{2/3} \left (\operatorname {AiryBi}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )+c_1 \operatorname {AiryAi}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )\right )} \\ y(x)\to \frac {b \operatorname {AiryAiPrime}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )}{(-a b)^{2/3} \operatorname {AiryAi}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )} \\ \end{align*}

2.1 problem 1

2.1.1 Solving as riccati ode

2.1.2 Maple step by step solution