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2.1.30 Problem 30

Solved using first_order_ode_riccati
✓Maple
✓Mathematica
✗Sympy

Internal problem ID [10288]
Book : First order enumerated odes
Section : section 1
Problem number : 30
Date solved : Thursday, November 27, 2025 at 10:30:58 AM
CAS classification : [_Riccati]

Solved using first_order_ode_riccati

Time used: 0.128 (sec)

Solve

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

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = b \,y^{2}+x +y \]
With Riccati ODE standard form
\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows that \(f_0(x)=x\), \(f_1(x)=1\) and \(f_2(x)=b\). Let
\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{b 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 &=b\\ f_2^2 f_0 &=x \,b^{2} \end{align*}

Substituting the above terms back in equation (2) gives

\[ b u^{\prime \prime }\left (x \right )-b u^{\prime }\left (x \right )+x \,b^{2} u \left (x \right ) = 0 \]
This is Airy ODE. It has the general form
\[ a \frac {d^{2}u}{d x^{2}} + b \frac {d u}{d x} + c u x = F(x) \]
Where in this case
\begin{align*} a &= b\\ b &= -b\\ c &= b^{2}\\ F &= 0 \end{align*}

Therefore the solution to the homogeneous Airy ODE becomes

\[ u = c_1 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (-\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )+c_2 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (-\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right ) \]
Taking derivative gives
\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (-\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )}{2}-c_1 \,{\mathrm e}^{\frac {x}{2}} b^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )+\frac {c_2 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (-\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )}{2}-c_2 \,{\mathrm e}^{\frac {x}{2}} b^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right ) \end{equation}
Substituting equations (3,4) into (1) results in
\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{b u} \\ y &= \frac {2 c_2 \,b^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {4 x b -1}{4 b^{{2}/{3}}}\right )+2 c_1 \,b^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {4 x b -1}{4 b^{{2}/{3}}}\right )-c_2 \operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )-c_1 \operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )}{2 b \left (c_2 \operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )+c_1 \operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )\right )} \\ \end{align*}
Doing change of constants, the above solution becomes
\[ y = -\frac {\frac {{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (-\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )}{2}-{\mathrm e}^{\frac {x}{2}} b^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )+\frac {c_3 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (-\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )}{2}-c_3 \,{\mathrm e}^{\frac {x}{2}} b^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )}{b \left ({\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (-\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )+c_3 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (-\frac {b^{3} x -\frac {1}{4} b^{2}}{b^{{8}/{3}}}\right )\right )} \]
Simplifying the above gives
\begin{align*} y &= \frac {2 c_3 \,b^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {4 x b -1}{4 b^{{2}/{3}}}\right )+2 b^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {4 x b -1}{4 b^{{2}/{3}}}\right )-c_3 \operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )-\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )}{2 b \left (c_3 \operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )+\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )\right )} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {2 c_3 \,b^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {4 x b -1}{4 b^{{2}/{3}}}\right )+2 b^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {4 x b -1}{4 b^{{2}/{3}}}\right )-c_3 \operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )-\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )}{2 b \left (c_3 \operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )+\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{{2}/{3}}}\right )\right )} \\ \end{align*}
✓ Maple. Time used: 0.007 (sec). Leaf size: 105
ode:=diff(y(x),x) = x+y(x)+b*y(x)^2; 
dsolve(ode,y(x), singsol=all);
\[ y = \frac {2 \operatorname {AiryAi}\left (1, -\frac {4 b x -1}{4 b^{{2}/{3}}}\right ) b^{{1}/{3}} c_1 -\operatorname {AiryAi}\left (-\frac {4 b x -1}{4 b^{{2}/{3}}}\right ) c_1 +2 \operatorname {AiryBi}\left (1, -\frac {4 b x -1}{4 b^{{2}/{3}}}\right ) b^{{1}/{3}}-\operatorname {AiryBi}\left (-\frac {4 b x -1}{4 b^{{2}/{3}}}\right )}{2 b \left (\operatorname {AiryAi}\left (-\frac {4 b x -1}{4 b^{{2}/{3}}}\right ) c_1 +\operatorname {AiryBi}\left (-\frac {4 b x -1}{4 b^{{2}/{3}}}\right )\right )} \]

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 sub-methods: 
   <- Abel AIR successful: ODE belongs to the 0F1 0-parameter (Airy type) class

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=x +y \left (x \right )+b y \left (x \right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=x +y \left (x \right )+b y \left (x \right )^{2} \end {array} \]
✓ Mathematica. Time used: 0.129 (sec). Leaf size: 211
ode=D[y[x],x]==x+y[x]+b*y[x]^2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to -\frac {-(-b)^{2/3} \operatorname {AiryBi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )+2 b \operatorname {AiryBiPrime}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )+c_1 \left (2 b \operatorname {AiryAiPrime}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )-(-b)^{2/3} \operatorname {AiryAi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )\right )}{2 (-b)^{5/3} \left (\operatorname {AiryBi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )+c_1 \operatorname {AiryAi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )\right )}\\ y(x)&\to -\frac {\frac {2 \sqrt [3]{-b} \operatorname {AiryAiPrime}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )}{\operatorname {AiryAi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )}+1}{2 b} \end{align*}
✗ Sympy
from sympy import * 
x = symbols("x") 
b = symbols("b") 
y = Function("y") 
ode = Eq(-b*y(x)**2 - x - y(x) + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

TypeError : bad operand type for unary -: list

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