2.1.30 Problem 30

Internal problem ID [9013]
Book : First order enumerated odes
Section : section 1
Problem number : 30
Date solved : Friday, April 25, 2025 at 05:34:19 PM
CAS classification : [_Riccati]

Solved using first_order_ode_riccati

Time used: 0.145 (sec)

Solve

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

Shows that $f_0(x)=x$, $f_1(x)=1$ and $f_2(x)=b$. Let

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

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

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

Substituting the above terms back in equation (2) gives

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

This is Airy ODE. It has the general form

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

Where in this case

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

Therefore the solution to the homogeneous Airy ODE becomes

$$u=c_1{\mathrm{e}}^{\frac{x}{2}}\mathrm{AiryAi}(-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})+c_2{\mathrm{e}}^{\frac{x}{2}}\mathrm{AiryBi}(-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})$$

Will add steps showing solving for IC soon.

Taking derivative gives

$$u^{\prime }\left(x\right)=\frac{c_1{\mathrm{e}}^{\frac{x}{2}}\mathrm{AiryAi}(-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})}{2}-c_1{\mathrm{e}}^{\frac{x}{2}}{b}^{1/3}\mathrm{AiryAi}(1,-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})+\frac{c_2{\mathrm{e}}^{\frac{x}{2}}\mathrm{AiryBi}(-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})}{2}-c_2{\mathrm{e}}^{\frac{x}{2}}{b}^{1/3}\mathrm{AiryBi}(1,-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})$$

Doing change of constants, the solution becomes

$$y=-\frac{\frac{c_3{\mathrm{e}}^{\frac{x}{2}}\mathrm{AiryAi}(-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})}{2}-c_3{\mathrm{e}}^{\frac{x}{2}}{b}^{1/3}\mathrm{AiryAi}(1,-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})+\frac{{\mathrm{e}}^{\frac{x}{2}}\mathrm{AiryBi}(-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})}{2}-{\mathrm{e}}^{\frac{x}{2}}{b}^{1/3}\mathrm{AiryBi}(1,-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})}{b(c_3{\mathrm{e}}^{\frac{x}{2}}\mathrm{AiryAi}(-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}})+{\mathrm{e}}^{\frac{x}{2}}\mathrm{AiryBi}(-\frac{b^{3}x-\frac{1}{4}{b}^2}{b^{8/3}}))}$$

Which simplifies to

$$\begin{array}{rl}y& =\frac{2c_3{b}^{1/3}\mathrm{AiryAi}(1,-\frac{4bx-1}{4b^{2/3}})+2b^{1/3}\mathrm{AiryBi}(1,-\frac{4bx-1}{4b^{2/3}})-c_3\mathrm{AiryAi}(-\frac{4bx-1}{4b^{2/3}})-\mathrm{AiryBi}(-\frac{4bx-1}{4b^{2/3}})}{2b(c_3\mathrm{AiryAi}(-\frac{4bx-1}{4b^{2/3}})+\mathrm{AiryBi}(-\frac{4bx-1}{4b^{2/3}}))}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{2c_3{b}^{1/3}\mathrm{AiryAi}(1,-\frac{4bx-1}{4b^{2/3}})+2b^{1/3}\mathrm{AiryBi}(1,-\frac{4bx-1}{4b^{2/3}})-c_3\mathrm{AiryAi}(-\frac{4bx-1}{4b^{2/3}})-\mathrm{AiryBi}(-\frac{4bx-1}{4b^{2/3}})}{2b(c_3\mathrm{AiryAi}(-\frac{4bx-1}{4b^{2/3}})+\mathrm{AiryBi}(-\frac{4bx-1}{4b^{2/3}}))}\end{array}$$

✓ Maple. Time used: 0.009 (sec). Leaf size: 105

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

$$y=\frac{2\mathrm{AiryAi}(1,-\frac{4bx-1}{4b^{2/3}})b^{1/3}{c}_1-\mathrm{AiryAi}(-\frac{4bx-1}{4b^{2/3}})c_1+2\mathrm{AiryBi}(1,-\frac{4bx-1}{4b^{2/3}})b^{1/3}-\mathrm{AiryBi}(-\frac{4bx-1}{4b^{2/3}})}{2b(\mathrm{AiryAi}(-\frac{4bx-1}{4b^{2/3}})c_1+\mathrm{AiryBi}(-\frac{4bx-1}{4b^{2/3}}))}$$

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

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

✓ Mathematica. Time used: 0.205 (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]
 

$$\begin{array}{r}y(x)\to -\frac{-(-b{)}^{2/3}\mathrm{AiryBi}\left(\frac{\frac{1}{4}-bx}{(-b{)}^{2/3}}\right)+2b\mathrm{AiryBiPrime}\left(\frac{\frac{1}{4}-bx}{(-b{)}^{2/3}}\right)+c_1(2b\mathrm{AiryAiPrime}\left(\frac{\frac{1}{4}-bx}{(-b{)}^{2/3}}\right)-(-b{)}^{2/3}\mathrm{AiryAi}\left(\frac{\frac{1}{4}-bx}{(-b{)}^{2/3}}\right))}{2(-b{)}^{5/3}(\mathrm{AiryBi}\left(\frac{\frac{1}{4}-bx}{(-b{)}^{2/3}}\right)+c_1\mathrm{AiryAi}\left(\frac{\frac{1}{4}-bx}{(-b{)}^{2/3}}\right))}\\ y(x)\to -\frac{\frac{2\sqrt[3]{-b}\mathrm{AiryAiPrime}\left(\frac{\frac{1}{4}-bx}{(-b{)}^{2/3}}\right)}{\mathrm{AiryAi}\left(\frac{\frac{1}{4}-bx}{(-b{)}^{2/3}}\right)}+1}{2b}\end{array}$$

✗ 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