2.1.37 Problem 38

Internal problem ID [8749]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 38
Date solved : Sunday, March 30, 2025 at 01:30:02 PM
CAS classification : [_rational, _Riccati]

Solved using first_order_ode_riccati

Time used: 0.219 (sec)

Solve

$$\begin{array}{rl}x{y}^{\prime }-y+y^2& =x^{2/3}\end{array}$$

In canonical form the ODE is

$$\begin{array}{rl}{y}^{\prime }& =F(x,y)\\ & =-\frac{y^2-x^{2/3}-y}{x}\end{array}$$

This is a Riccati ODE. Comparing the ODE to solve

$$y^{\prime }=-\frac{y^2}{x}+\frac{1}{x^{1/3}}+\frac{y}{x}$$

With Riccati ODE standard form

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

Shows that $f_0(x)=\frac{1}{x^{1/3}}$, $f_1(x)=\frac{1}{x}$ and $f_2(x)=-\frac{1}{x}$. Let

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

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

$$\begin{array}{rl}{f}_2^{\prime }& =\frac{1}{x^2}\\ f_1{f}_2& =-\frac{1}{x^2}\\ f_2^2{f}_0& =\frac{1}{x^{7/3}}\end{array}$$

Substituting the above terms back in equation (2) gives

$$\begin{array}{r}-\frac{u^{\prime \prime }\left(x\right)}{x}+\frac{u\left(x\right)}{x^{7/3}}=0\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& x^2{u}^{\prime \prime }-x^{2/3}u=0\end{array}$$

Bessel ode has the form

$$\begin{array}{}\text{(2)}& x^2{u}^{\prime \prime }+u^{\prime }x+(-n^2+x^2)u=0\end{array}$$

The generalized form of Bessel ode is given by Bowman (1958) as the following

$$\begin{array}{}\text{(3)}& x^2{u}^{\prime \prime }+(1-2\alpha )x{u}^{\prime }+({\beta }^2{\gamma }^2{x}^{2\gamma }-n^2{\gamma }^2+{\alpha }^2)u=0\end{array}$$

With the standard solution

$$\begin{array}{}\text{(4)}& u& =x^{\alpha }(c_1\mathrm{BesselJ}(n,\beta x^{\gamma })+c_2\mathrm{BesselY}(n,\beta x^{\gamma }))\end{array}$$

Comparing (3) to (1) and solving for $\alpha ,\beta ,n,\gamma$ gives

$$\begin{array}{rl}\alpha & =\frac{1}{2}\\ \beta & =3i\\ n& =\frac{3}{2}\\ \gamma & =\frac{1}{3}\end{array}$$

Substituting all the above into (4) gives the solution as

$$\begin{array}{r}u=-\frac{c_1{x}^{1/6}\sqrt{2}\sqrt{3}(3\cosh \left(3x^{1/3}\right)x^{1/3}-\sinh \left(3x^{1/3}\right))}{9\sqrt{\pi }\sqrt{i{x}^{1/3}}}+\frac{i{c}_2{x}^{1/6}\sqrt{2}\sqrt{3}(-3\sinh \left(3x^{1/3}\right)x^{1/3}+\cosh \left(3x^{1/3}\right))}{9\sqrt{\pi }\sqrt{i{x}^{1/3}}}\end{array}$$

Will add steps showing solving for IC soon.

Taking derivative gives

$$u^{\prime }\left(x\right)=-\frac{c_1\sqrt{2}\sqrt{3}(3\cosh \left(3x^{1/3}\right)x^{1/3}-\sinh \left(3x^{1/3}\right))}{54x^{5/6}\sqrt{\pi }\sqrt{i{x}^{1/3}}}-\frac{c_1\sqrt{2}\sqrt{3}\sinh \left(3x^{1/3}\right)}{3x^{1/6}\sqrt{\pi }\sqrt{i{x}^{1/3}}}+\frac{i{c}_1\sqrt{2}\sqrt{3}(3\cosh \left(3x^{1/3}\right)x^{1/3}-\sinh \left(3x^{1/3}\right))}{54\sqrt{x}\sqrt{\pi }{\left(i{x}^{1/3}\right)}^{3/2}}+\frac{i{c}_2\sqrt{2}\sqrt{3}(-3\sinh \left(3x^{1/3}\right)x^{1/3}+\cosh \left(3x^{1/3}\right))}{54x^{5/6}\sqrt{\pi }\sqrt{i{x}^{1/3}}}-\frac{i{c}_2\sqrt{2}\sqrt{3}\cosh \left(3x^{1/3}\right)}{3x^{1/6}\sqrt{\pi }\sqrt{i{x}^{1/3}}}+\frac{c_2\sqrt{2}\sqrt{3}(-3\sinh \left(3x^{1/3}\right)x^{1/3}+\cosh \left(3x^{1/3}\right))}{54\sqrt{x}\sqrt{\pi }{\left(i{x}^{1/3}\right)}^{3/2}}$$

Doing change of constants, the solution becomes

$$y=\frac{(-\frac{c_3\sqrt{2}\sqrt{3}(3\cosh \left(3x^{1/3}\right)x^{1/3}-\sinh \left(3x^{1/3}\right))}{54x^{5/6}\sqrt{\pi }\sqrt{i{x}^{1/3}}}-\frac{c_3\sqrt{2}\sqrt{3}\sinh \left(3x^{1/3}\right)}{3x^{1/6}\sqrt{\pi }\sqrt{i{x}^{1/3}}}+\frac{i{c}_3\sqrt{2}\sqrt{3}(3\cosh \left(3x^{1/3}\right)x^{1/3}-\sinh \left(3x^{1/3}\right))}{54\sqrt{x}\sqrt{\pi }{\left(i{x}^{1/3}\right)}^{3/2}}+\frac{i\sqrt{2}\sqrt{3}(-3\sinh \left(3x^{1/3}\right)x^{1/3}+\cosh \left(3x^{1/3}\right))}{54x^{5/6}\sqrt{\pi }\sqrt{i{x}^{1/3}}}-\frac{i\sqrt{2}\sqrt{3}\cosh \left(3x^{1/3}\right)}{3x^{1/6}\sqrt{\pi }\sqrt{i{x}^{1/3}}}+\frac{\sqrt{2}\sqrt{3}(-3\sinh \left(3x^{1/3}\right)x^{1/3}+\cosh \left(3x^{1/3}\right))}{54\sqrt{x}\sqrt{\pi }{\left(i{x}^{1/3}\right)}^{3/2}})x}{-\frac{c_3{x}^{1/6}\sqrt{2}\sqrt{3}(3\cosh \left(3x^{1/3}\right)x^{1/3}-\sinh \left(3x^{1/3}\right))}{9\sqrt{\pi }\sqrt{i{x}^{1/3}}}+\frac{i{x}^{1/6}\sqrt{2}\sqrt{3}(-3\sinh \left(3x^{1/3}\right)x^{1/3}+\cosh \left(3x^{1/3}\right))}{9\sqrt{\pi }\sqrt{i{x}^{1/3}}}}$$

Which simplifies to

$$\begin{array}{rl}y& =\frac{3x^{2/3}(i\sinh \left(3x^{1/3}\right)c_3-\cosh \left(3x^{1/3}\right))}{(-i{c}_3-3x^{1/3})\sinh \left(3x^{1/3}\right)+(3i{x}^{1/3}{c}_3+1)\cosh \left(3x^{1/3}\right)}\end{array}$$

Figure 2.76: Slope field $x{y}^{\prime }-y+y^2=x^{2/3}$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{3x^{2/3}(i\sinh \left(3x^{1/3}\right)c_3-\cosh \left(3x^{1/3}\right))}{(-i{c}_3-3x^{1/3})\sinh \left(3x^{1/3}\right)+(3i{x}^{1/3}{c}_3+1)\cosh \left(3x^{1/3}\right)}\end{array}$$

✓ Maple. Time used: 0.004 (sec). Leaf size: 72

ode:=diff(y(x),x)*x-y(x)+y(x)^2 = x^(2/3); 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{(c_1|3x^{1/3}-1|{\mathrm{e}}^{6x^{1/3}}+\mathrm{abs}(1,3x^{1/3}-1){\mathrm{e}}^{6x^{1/3}}{c}_1-3x^{1/3})x^{1/3}}{c_1|3x^{1/3}-1|{\mathrm{e}}^{6x^{1/3}}+3x^{1/3}+1}$$

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: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = 1/x^(4/3)*y(x), y(x) 
      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Trying an equivalence, under non-integer power transformations, 
         to LODEs admitting Liouvillian solutions. 
         -> Trying a Liouvillian solution using Kovacics algorithm 
            A Liouvillian solution exists 
            Group is reducible or imprimitive 
         <- Kovacics algorithm successful 
      <- Equivalence, under non-integer power transformations successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x\left(\frac{d}{dx}y\left(x\right)\right)-y\left(x\right)+y{\left(x\right)}^2=x^{2/3}\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=\frac{y\left(x\right)-y{\left(x\right)}^2+x^{2/3}}{x}\end{array}$$

✓ Mathematica. Time used: 0.2 (sec). Leaf size: 131

ode=x*D[y[x],x]-y[x]+y[x]^2==x^(2/3); 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}y(x)\to \frac{3x^{2/3}(c_1\cosh \left(3\sqrt[3]{x}\right)-i\sinh \left(3\sqrt[3]{x}\right))}{(-3i\sqrt[3]{x}-c_1)\cosh \left(3\sqrt[3]{x}\right)+(3c_1\sqrt[3]{x}+i)\sinh \left(3\sqrt[3]{x}\right)}\\ y(x)\to \frac{3x^{2/3}\cosh \left(3\sqrt[3]{x}\right)}{3\sqrt[3]{x}\sinh \left(3\sqrt[3]{x}\right)-\cosh \left(3\sqrt[3]{x}\right)}\end{array}$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x**(2/3) + x*Derivative(y(x), x) + y(x)**2 - y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE Derivative(y(x), x) + y(x)**2/x - y(x)/x - 1/x**(1/3) cannot be solved by the factorable group method