2.1.19 Problem 19
Internal
problem
ID
[10277]
Book
:
First
order
enumerated
odes
Section
:
section
1
Problem
number
:
19
Date
solved
:
Thursday, November 27, 2025 at 10:29:36 AM
CAS
classification
:
[[_Riccati, _special]]
Solved using first_order_ode_reduced_riccati
Time used: 0.069 (sec)
Solve
\begin{align*}
c y^{\prime }&=a x +b y^{2} \\
\end{align*}
This is reduced Riccati ode of the form
\begin{align*} y^{\prime }&=a \,x^{n}+b y^{2} \end{align*}
Comparing the given ode to the above shows that
\begin{align*} a &= \frac {a}{c}\\ b &= \frac {b}{c}\\ n &= 1 \end{align*}
Since \(n\neq -2\) then the solution of the reduced Riccati ode is given by
\begin{align*} w & =\sqrt {x}\left \{ \begin {array}[c]{cc} c_{1}\operatorname {BesselJ}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab} x^{k}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) & ab>0\\ c_{1}\operatorname {BesselI}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-ab}x^{k}\right ) +c_{2}\operatorname {BesselK}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-ab}x^{k}\right ) & ab<0 \end {array} \right . \tag {1}\\ y & =-\frac {1}{b}\frac {w^{\prime }}{w}\nonumber \\ k &=1+\frac {n}{2}\nonumber \end{align*}
EQ(1) gives
\begin{align*} k &= {\frac {3}{2}}\\ w &= \sqrt {x}\, \left (c_1 \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )+c_2 \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )\right ) \end{align*}
Therefore the solution becomes
\begin{align*} y & =-\frac {1}{b}\frac {w^{\prime }}{w} \end{align*}
Substituting the value of \(b,w\) found above and simplifying gives
\[
y = \frac {\left (-\operatorname {BesselJ}\left (-\frac {2}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselY}\left (-\frac {2}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right ) c_2 \right ) c \sqrt {\frac {a b}{c^{2}}}\, \sqrt {x}}{b \left (c_1 \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )+c_2 \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )\right )}
\]
Letting
\(c_2 = 1\) the above becomes
\[
y = \frac {\left (-\operatorname {BesselY}\left (-\frac {2}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )-\operatorname {BesselJ}\left (-\frac {2}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right ) c_1 \right ) c \sqrt {\frac {a b}{c^{2}}}\, \sqrt {x}}{b \left (c_1 \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )+\operatorname {BesselY}\left (\frac {1}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )\right )}
\]
Summary of solutions found
\begin{align*}
y &= \frac {\left (-\operatorname {BesselY}\left (-\frac {2}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )-\operatorname {BesselJ}\left (-\frac {2}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right ) c_1 \right ) c \sqrt {\frac {a b}{c^{2}}}\, \sqrt {x}}{b \left (c_1 \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )+\operatorname {BesselY}\left (\frac {1}{3}, \frac {2 \sqrt {\frac {a b}{c^{2}}}\, x^{{3}/{2}}}{3}\right )\right )} \\
\end{align*}
Solved using first_order_ode_riccati
Time used: 0.111 (sec)
Solve
\begin{align*}
c y^{\prime }&=a x +b y^{2} \\
\end{align*}
In canonical form the ODE is
\begin{align*} y' &= F(x,y)\\ &= \frac {b \,y^{2}+a x}{c} \end{align*}
This is a Riccati ODE. Comparing the ODE to solve
\[
y' = \frac {b \,y^{2}}{c}+\frac {a x}{c}
\]
With Riccati ODE standard form
\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows
that
\(f_0(x)=\frac {a x}{c}\),
\(f_1(x)=0\) and
\(f_2(x)=\frac {b}{c}\). Let
\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {b u}{c}} \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 &=\frac {b^{2} a x}{c^{3}} \end{align*}
Substituting the above terms back in equation (2) gives
\[
\frac {b u^{\prime \prime }\left (x \right )}{c}+\frac {b^{2} a x u \left (x \right )}{c^{3}} = 0
\]
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 &= \frac {b}{c}\\ b &= 0\\ c &= \frac {b^{2} a}{c^{3}}\\ F &= 0 \end{align*}
Therefore the solution to the homogeneous Airy ODE becomes
\[
u = c_1 \operatorname {AiryAi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )+c_2 \operatorname {AiryBi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )
\]
Taking derivative gives
\begin{equation}
\tag{4} u^{\prime }\left (x \right ) = -c_1 \left (\frac {a b}{c^{2}}\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )-c_2 \left (\frac {a b}{c^{2}}\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )
\end{equation}
Substituting equations (3,4) into (1) results in
\begin{align*}
y &= \frac {-u'}{f_2 u} \\
y &= \frac {-u'}{\frac {b u}{c}} \\
y &= \frac {\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} \left (\operatorname {AiryBi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right ) c_2 +\operatorname {AiryAi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right ) c_1 \right ) c}{b \left (c_1 \operatorname {AiryAi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )+c_2 \operatorname {AiryBi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )\right )} \\
\end{align*}
Doing change of constants, the above solution
becomes
\[
y = -\frac {\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )-c_3 \left (\frac {a b}{c^{2}}\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )\right ) c}{b \left (\operatorname {AiryAi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )+c_3 \operatorname {AiryBi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )\right )}
\]
Simplifying the above gives
\begin{align*}
y &= \frac {\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} \left (\operatorname {AiryBi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right ) c_3 +\operatorname {AiryAi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )\right ) c}{b \left (\operatorname {AiryAi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )+c_3 \operatorname {AiryBi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )\right )} \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= \frac {\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} \left (\operatorname {AiryBi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right ) c_3 +\operatorname {AiryAi}\left (1, -\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )\right ) c}{b \left (\operatorname {AiryAi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )+c_3 \operatorname {AiryBi}\left (-\left (\frac {a b}{c^{2}}\right )^{{1}/{3}} x \right )\right )} \\
\end{align*}
✓ Maple. Time used: 0.000 (sec). Leaf size: 75
ode:=c*diff(y(x),x) = a*x+b*y(x)^2;
dsolve(ode,y(x), singsol=all);
\[
y = \frac {\left (\frac {b a}{c^{2}}\right )^{{1}/{3}} \left (\operatorname {AiryAi}\left (1, -\left (\frac {b a}{c^{2}}\right )^{{1}/{3}} x \right ) c_1 +\operatorname {AiryBi}\left (1, -\left (\frac {b a}{c^{2}}\right )^{{1}/{3}} x \right )\right ) c}{b \left (c_1 \operatorname {AiryAi}\left (-\left (\frac {b a}{c^{2}}\right )^{{1}/{3}} x \right )+\operatorname {AiryBi}\left (-\left (\frac {b a}{c^{2}}\right )^{{1}/{3}} x \right )\right )}
\]
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 Special
<- Riccati Special successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & c \left (\frac {d}{d x}y \left (x \right )\right )=a x +b y \left (x \right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {a x +b y \left (x \right )^{2}}{c} \end {array} \]
✓ Mathematica. Time used: 0.122 (sec). Leaf size: 437
ode=c*D[y[x],x]==a*x+b*y[x]^2;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to \frac {c \left (x^{3/2} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} \left (-2 \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )+c_1 \left (\operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )-\operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )\right )}{2 b x \left (\operatorname {BesselJ}\left (\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )\right )}\\ y(x)&\to -\frac {c \left (x^{3/2} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )-x^{3/2} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )\right )}{2 b x \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c}} \sqrt {\frac {b}{c}} x^{3/2}\right )} \end{align*}
✗ Sympy
from sympy import *
x = symbols("x")
a = symbols("a")
b = symbols("b")
c = symbols("c")
y = Function("y")
ode = Eq(-a*x - b*y(x)**2 + c*Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
RecursionError : maximum recursion depth exceeded