2.2.41 Problem 41

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

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

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

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

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

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

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

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

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

$$y=\frac{\sqrt{x}(\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1-\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3})c_2)}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+c_2\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}$$

Letting $c_2=1$ the above becomes

$$y=\frac{\sqrt{x}(\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1-\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}$$

Figure 2.126: Slope field $y^{\prime }=x-y^2$

Summary of solutions found

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

Solved using first_order_ode_riccati

Time used: 0.210 (sec)

Solve

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

Shows that $f_0(x)=x$, $f_1(x)=0$ and $f_2(x)=-1$. Let

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

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

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

Substituting the above terms back in equation (2) gives

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

This is Airy ODE. It has the general form

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

Where in this case

$$\begin{array}{rl}a& =-1\\ b& =0\\ c& =1\\ F& =0\end{array}$$

Therefore the solution to the homogeneous Airy ODE becomes

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

Will add steps showing solving for IC soon.

Taking derivative gives

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

Doing change of constants, the solution becomes

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

Which simplifies to

$$\begin{array}{rl}y& =-\frac{(1+i\sqrt{3})(\mathrm{AiryAi}(1,-\frac{x(1+i\sqrt{3})}{2})c_5+\mathrm{AiryBi}(1,-\frac{x(1+i\sqrt{3})}{2}))}{2c_5\mathrm{AiryAi}(-\frac{x(1+i\sqrt{3})}{2})+2\mathrm{AiryBi}(-\frac{x(1+i\sqrt{3})}{2})}\end{array}$$

Figure 2.127: Slope field $y^{\prime }=x-y^2$

Summary of solutions found

$$\begin{array}{rl}y& =-\frac{(1+i\sqrt{3})(\mathrm{AiryAi}(1,-\frac{x(1+i\sqrt{3})}{2})c_5+\mathrm{AiryBi}(1,-\frac{x(1+i\sqrt{3})}{2}))}{2c_5\mathrm{AiryAi}(-\frac{x(1+i\sqrt{3})}{2})+2\mathrm{AiryBi}(-\frac{x(1+i\sqrt{3})}{2})}\end{array}$$

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

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

$$y=\frac{c_1\mathrm{AiryAi}(1,x)+\mathrm{AiryBi}(1,x)}{c_1\mathrm{AiryAi}\left(x\right)+\mathrm{AiryBi}\left(x\right)}$$

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

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

✓ 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]
 

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

✗ 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