Problem 41

2.2.41 Problem 41

Solved using ﬁrst_order_ode_reduced_riccati
Solved using ﬁrst_order_ode_riccati
✓Maple
✓Mathematica
✗Sympy

Internal problem ID [9164]
Book : Second order enumerated odes
Section : section 2
Problem number : 41
Date solved : Sunday, March 30, 2025 at 02:24:33 PM
CAS classiﬁcation : [[_Riccati, _special]]

Solved using ﬁrst_order_ode_reduced_riccati

Time used: 0.056 (sec)

Solve

\begin{align*} y^{\prime }&=x -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 &= 1\\ b &= -1\\ 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*}

Since \(ab<0\) then EQ(1) gives

\begin{align*} k &= {\frac {3}{2}}\\ w &= \sqrt {x}\, \left (c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+c_2 \operatorname {BesselK}\left (\frac {1}{3}, \frac {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 simplyﬁng gives

\[ y = \frac {\sqrt {x}\, \left (\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_2 \right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+c_2 \operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )} \]

Letting \(c_2 = 1\) the above becomes

\[ y = \frac {\sqrt {x}\, \left (\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )} \]

pict — Figure 2.126: Slope ﬁeld \(y^{\prime } = x -y^{2}\)

Summary of solutions found

\begin{align*} y &= \frac {\sqrt {x}\, \left (\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )} \\ \end{align*}

Solved using ﬁrst_order_ode_riccati

Time used: 0.210 (sec)

Solve

\begin{align*} y^{\prime }&=x -y^{2} \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= -y^{2}+x \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = -y^{2}+x \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=x\), \(f_1(x)=0\) 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 &=0\\ f_2^2 f_0 &=x \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} -u^{\prime \prime }\left (x \right )+x u \left (x \right ) = 0 \end{align*}

This is Airy ODE. It has the general form

\[ a u^{\prime \prime } + b u^{\prime } + c x u = F(x) \]

Where in this case

\begin{align*} a &= -1\\ b &= 0\\ c &= 1\\ F &= 0 \end{align*}

Therefore the solution to the homogeneous Airy ODE becomes

\[ u = c_3 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_4 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right ) \]

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = -c_3 \left (-1\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )-c_4 \left (-1\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -x \left (-1\right )^{{1}/{3}}\right ) \]

Doing change of constants, the solution becomes

\[ y = \frac {-c_5 \left (-1\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )-\left (-1\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )}{c_5 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+\operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right )} \]

Which simpliﬁes to

\begin{align*} y &= -\frac {\left (1+i \sqrt {3}\right ) \left (\operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right ) c_5 +\operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{2 c_5 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+2 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )} \\ \end{align*}

Summary of solutions found

✓ Maple. Time used: 0.001 (sec). Leaf size: 23

ode:=diff(y(x),x) = x-y(x)^2; 
dsolve(ode,y(x), singsol=all);

\[ y = \frac {c_1 \operatorname {AiryAi}\left (1, x\right )+\operatorname {AiryBi}\left (1, x\right )}{c_1 \operatorname {AiryAi}\left (x \right )+\operatorname {AiryBi}\left (x \right )} \]

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

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

✓ Mathematica. Time used: 0.125 (sec). Leaf size: 223

ode=D[y[x],x]==x-y[x]^2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]

\begin{align*} y(x)\to -\frac {-i x^{3/2} \left (2 \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2}{3} i x^{3/2}\right )+c_1 \left (\operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} i x^{3/2}\right )-\operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} i x^{3/2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )}{2 x \left (\operatorname {BesselJ}\left (\frac {1}{3},\frac {2}{3} i x^{3/2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )\right )} \\ y(x)\to \frac {i x^{3/2} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} i x^{3/2}\right )-i x^{3/2} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} i x^{3/2}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )}{2 x \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )} \\ \end{align*}

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x + y(x)**2 + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)

TypeError : bad operand type for unary -: list

[next] [prev] [prev-tail] [front] [up]