2.1.11 Problem 11

Internal problem ID [8994]
Book : First order enumerated odes
Section : section 1
Problem number : 11
Date solved : Friday, April 18, 2025 at 01:35:46 PM
CAS classification : [[_Riccati, _special]]

Solved using first_order_ode_reduced_riccati

Time used: 0.137 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }& =ax+b{y}^2\end{array}$$

This is reduced Riccati ode of the form

$$\begin{array}{rl}{y}^{\prime }& =a{x}^n+b{y}^2\end{array}$$

Comparing the given ode to the above shows that

$$\begin{array}{rl}a& =a\\ b& =b\\ n& =1\end{array}$$

Since $n\ne -2$ then the solution of the reduced Riccati ode is given by

$$\begin{array}{rl}\text{(1)}& w& =\sqrt{x}\{\begin{array}{cc}{c}_1\mathrm{BesselJ}(\frac{1}{2k},\frac{1}{k}\sqrt{ab}{x}^k)+c_2\mathrm{BesselY}(\frac{1}{2k},\frac{1}{k}\sqrt{ab}{x}^k)& ab>0\\ c_1\mathrm{BesselI}(\frac{1}{2k},\frac{1}{k}\sqrt{-ab}{x}^k)+c_2\mathrm{BesselK}(\frac{1}{2k},\frac{1}{k}\sqrt{-ab}{x}^k)& ab<0\end{array}y& =-\frac{1}{b}\frac{w^{\prime }}{w}\\ k& =1+\frac{n}{2}\end{array}$$

EQ(1) gives

$$\begin{array}{rl}k& =\frac{3}{2}\\ w& =\sqrt{x}(c_1\mathrm{BesselJ}(\frac{1}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})+c_2\mathrm{BesselY}(\frac{1}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3}))\end{array}$$

Therefore the solution becomes

$$\begin{array}{rl}y& =-\frac{1}{b}\frac{w^{\prime }}{w}\end{array}$$

Substituting the value of $b,w$ found above and simplyfing gives

$$y=-\frac{\sqrt{ab}\sqrt{x}(\mathrm{BesselY}(-\frac{2}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})c_2+\mathrm{BesselJ}(-\frac{2}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})c_1)}{b(c_1\mathrm{BesselJ}(\frac{1}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})+c_2\mathrm{BesselY}(\frac{1}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3}))}$$

Letting $c_2=1$ the above becomes

$$y=-\frac{\sqrt{ab}\sqrt{x}(\mathrm{BesselY}(-\frac{2}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})+\mathrm{BesselJ}(-\frac{2}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})c_1)}{b(c_1\mathrm{BesselJ}(\frac{1}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})+\mathrm{BesselY}(\frac{1}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3}))}$$

Summary of solutions found

$$\begin{array}{rl}y& =-\frac{\sqrt{ab}\sqrt{x}(\mathrm{BesselY}(-\frac{2}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})+\mathrm{BesselJ}(-\frac{2}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})c_1)}{b(c_1\mathrm{BesselJ}(\frac{1}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3})+\mathrm{BesselY}(\frac{1}{3},\frac{2\sqrt{ab}{x}^{3/2}}{3}))}\end{array}$$

Solved using first_order_ode_riccati

Time used: 0.168 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }& =ax+b{y}^2\end{array}$$

In canonical form the ODE is

$$\begin{array}{rl}{y}^{\prime }& =F(x,y)\\ & =b{y}^2+ax\end{array}$$

This is a Riccati ODE. Comparing the ODE to solve

$$y^{\prime }=b{y}^2+ax$$

With Riccati ODE standard form

$$y^{\prime }=f_0(x)+f_1(x)y+f_2(x)y^2$$

Shows that $f_0(x)=ax$, $f_1(x)=0$ and $f_2(x)=b$. Let

$$\begin{array}{rl}y& =\frac{-u^{\prime }}{f_{2}u}\\ \text{(1)}& & =\frac{-u^{\prime }}{ub}\end{array}$$

Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for $u(x)$ which is

$$\begin{array}{}\text{(2)}& f_2{u}^{″}(x)-(f_2^{\prime }+f_1{f}_2)u^{\prime }(x)+f_2^2{f}_{0}u(x)& =0\end{array}$$

But

$$\begin{array}{rl}{f}_2^{\prime }& =0\\ f_1{f}_2& =0\\ f_2^2{f}_0& =b^{2}ax\end{array}$$

Substituting the above terms back in equation (2) gives

$$\begin{array}{r}b{u}^{\prime \prime }\left(x\right)+b^{2}axu\left(x\right)=0\end{array}$$

This is Airy ODE. It has the general form

$$a{u}^{\prime \prime }+b{u}^{\prime }+cux=F(x)$$

Where in this case

$$\begin{array}{rl}a& =b\\ b& =0\\ c& =b^{2}a\\ F& =0\end{array}$$

Therefore the solution to the homogeneous Airy ODE becomes

$$u=c_3\mathrm{AiryAi}(-{\left(ab\right)}^{1/3}x)+c_4\mathrm{AiryBi}(-{\left(ab\right)}^{1/3}x)$$

Will add steps showing solving for IC soon.

Taking derivative gives

$$u^{\prime }\left(x\right)=-c_3{\left(ab\right)}^{1/3}\mathrm{AiryAi}(1,-{\left(ab\right)}^{1/3}x)-c_4{\left(ab\right)}^{1/3}\mathrm{AiryBi}(1,-{\left(ab\right)}^{1/3}x)$$

Doing change of constants, the solution becomes

$$y=-\frac{-c_5{\left(ab\right)}^{1/3}\mathrm{AiryAi}(1,-{\left(ab\right)}^{1/3}x)-{\left(ab\right)}^{1/3}\mathrm{AiryBi}(1,-{\left(ab\right)}^{1/3}x)}{b(c_5\mathrm{AiryAi}(-{\left(ab\right)}^{1/3}x)+\mathrm{AiryBi}(-{\left(ab\right)}^{1/3}x))}$$

Which simplifies to

$$\begin{array}{rl}y& =\frac{{\left(ab\right)}^{1/3}(\mathrm{AiryAi}(1,-{\left(ab\right)}^{1/3}x)c_5+\mathrm{AiryBi}(1,-{\left(ab\right)}^{1/3}x))}{b(c_5\mathrm{AiryAi}(-{\left(ab\right)}^{1/3}x)+\mathrm{AiryBi}(-{\left(ab\right)}^{1/3}x))}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{{\left(ab\right)}^{1/3}(\mathrm{AiryAi}(1,-{\left(ab\right)}^{1/3}x)c_5+\mathrm{AiryBi}(1,-{\left(ab\right)}^{1/3}x))}{b(c_5\mathrm{AiryAi}(-{\left(ab\right)}^{1/3}x)+\mathrm{AiryBi}(-{\left(ab\right)}^{1/3}x))}\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 59

ode:=diff(y(x),x) = a*x+b*y(x)^2; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{{\left(ba\right)}^{1/3}(\mathrm{AiryAi}(1,-{\left(ba\right)}^{1/3}x)c_1+\mathrm{AiryBi}(1,-{\left(ba\right)}^{1/3}x))}{b(c_1\mathrm{AiryAi}(-{\left(ba\right)}^{1/3}x)+\mathrm{AiryBi}(-{\left(ba\right)}^{1/3}x))}$$

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}y\left(x\right)=ax+by{\left(x\right)}^2\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=ax+by{\left(x\right)}^2\end{array}$$

✓ Mathematica. Time used: 0.165 (sec). Leaf size: 331

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

$$\begin{array}{r}y(x)\to \frac{\sqrt{a}\sqrt{b}{x}^{3/2}(-2\mathrm{BesselJ}(-\frac{2}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2})+c_1(\mathrm{BesselJ}(\frac{2}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2})-\mathrm{BesselJ}(-\frac{4}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2})))-c_1\mathrm{BesselJ}(-\frac{1}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2})}{2bx(\mathrm{BesselJ}(\frac{1}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2})+c_1\mathrm{BesselJ}(-\frac{1}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2}))}\\ y(x)\to -\frac{\sqrt{a}\sqrt{b}{x}^{3/2}\mathrm{BesselJ}(-\frac{4}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2})-\sqrt{a}\sqrt{b}{x}^{3/2}\mathrm{BesselJ}(\frac{2}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2})+\mathrm{BesselJ}(-\frac{1}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2})}{2bx\mathrm{BesselJ}(-\frac{1}{3},\frac{2}{3}\sqrt{a}\sqrt{b}{x}^{3/2})}\end{array}$$

✗ Sympy

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

NotImplementedError : The given ODE -a*x - b*y(x)**2 + Derivative(y(x), x) cannot be solved by the lie group method