2.1.85 Problem 84

Internal problem ID [8797]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 84
Date solved : Sunday, March 30, 2025 at 01:37:36 PM
CAS classification : [_Riccati]

Solved using first_order_ode_riccati

Time used: 0.791 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }& =x^2+y^2-1\end{array}$$

In canonical form the ODE is

$$\begin{array}{rl}{y}^{\prime }& =F(x,y)\\ & =x^2+y^2-1\end{array}$$

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

Shows that $f_0(x)=x^2-1$, $f_1(x)=0$ 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

$$\begin{array}{rl}{f}_2^{\prime }& =0\\ f_1{f}_2& =0\\ f_2^2{f}_0& =x^2-1\end{array}$$

Substituting the above terms back in equation (2) gives

$$\begin{array}{r}{u}^{\prime \prime }\left(x\right)+(x^2-1)u\left(x\right)=0\end{array}$$

Unable to solve. Will ask Maple to solve this ode now.

Solution obtained is

$$u\left(x\right)=\frac{c_1\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+c_2\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)}{\sqrt{x}}$$

Taking derivative gives

$$u^{\prime }\left(x\right)=\frac{2i{c}_1((\frac{1}{2}-\frac{1}{4x^2})\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+\frac{(\frac{1}{4}-\frac{3i}{4})\mathrm{WhittakerM}(1+\frac{i}{4},\frac{1}{4},i{x}^2)}{x^2})x+2i{c}_2((\frac{1}{2}-\frac{1}{4x^2})\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)+\frac{i\mathrm{WhittakerW}(1+\frac{i}{4},\frac{1}{4},i{x}^2)}{x^2})x}{\sqrt{x}}-\frac{c_1\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+c_2\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)}{2x^{3/2}}$$

Doing change of constants, the solution becomes

$$y=-\frac{(\frac{2i{c}_5((\frac{1}{2}-\frac{1}{4x^2})\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+\frac{(\frac{1}{4}-\frac{3i}{4})\mathrm{WhittakerM}(1+\frac{i}{4},\frac{1}{4},i{x}^2)}{x^2})x+2i((\frac{1}{2}-\frac{1}{4x^2})\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)+\frac{i\mathrm{WhittakerW}(1+\frac{i}{4},\frac{1}{4},i{x}^2)}{x^2})x}{\sqrt{x}}-\frac{c_5\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)}{2x^{3/2}})\sqrt{x}}{c_5\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)}$$

Which simplifies to

$$\begin{array}{rl}y& =\frac{(-3-i)c_5\mathrm{WhittakerM}(1+\frac{i}{4},\frac{1}{4},i{x}^2)+4\mathrm{WhittakerW}(1+\frac{i}{4},\frac{1}{4},i{x}^2)+(-2i{x}^2+i+1)c_5\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+(-2i{x}^2+i+1)\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)}{2x(c_5\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2))}\end{array}$$

Figure 2.154: Slope field $y^{\prime }=x^2+y^2-1$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{(-3-i)c_5\mathrm{WhittakerM}(1+\frac{i}{4},\frac{1}{4},i{x}^2)+4\mathrm{WhittakerW}(1+\frac{i}{4},\frac{1}{4},i{x}^2)+(-2i{x}^2+i+1)c_5\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+(-2i{x}^2+i+1)\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)}{2x(c_5\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2))}\end{array}$$

✓ Maple. Time used: 0.003 (sec). Leaf size: 85

ode:=diff(y(x),x) = x^2+y(x)^2-1; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{(-3-i)\mathrm{WhittakerM}(1+\frac{i}{4},\frac{1}{4},i{x}^2)+4\mathrm{WhittakerW}(1+\frac{i}{4},\frac{1}{4},i{x}^2)c_1+(-2i{x}^2+i+1)\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2)+(-2i{x}^2+i+1)c_1\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)}{2x(c_1\mathrm{WhittakerW}(\frac{i}{4},\frac{1}{4},i{x}^2)+\mathrm{WhittakerM}(\frac{i}{4},\frac{1}{4},i{x}^2))}$$

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 
trying Riccati sub-methods: 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = (-x^2+1)*y(x), y(x) 
      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a quadrature 
      checking if the LODE has constant coefficients 
      checking if the LODE is of Euler type 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Trying a Liouvillian solution using Kovacics algorithm 
      <- No Liouvillian solutions exists 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         -> elliptic 
         -> Legendre 
         -> Whittaker 
            -> hyper3: Equivalence to 1F1 under a power @ Moebius 
            <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
         <- Whittaker successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

✓ Mathematica. Time used: 0.22 (sec). Leaf size: 153

ode=D[y[x],x]==x^2+y[x]^2-1; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}y(x)\to \frac{i(x\mathrm{ParabolicCylinderD}(-\frac{1}{2}-\frac{i}{2},(-1+i)x)+(1+i)\mathrm{ParabolicCylinderD}(\frac{1}{2}-\frac{i}{2},(-1+i)x)-c_{1}x\mathrm{ParabolicCylinderD}(-\frac{1}{2}+\frac{i}{2},(1+i)x)+(1-i)c_1\mathrm{ParabolicCylinderD}(\frac{1}{2}+\frac{i}{2},(1+i)x))}{\mathrm{ParabolicCylinderD}(-\frac{1}{2}-\frac{i}{2},(-1+i)x)+c_1\mathrm{ParabolicCylinderD}(-\frac{1}{2}+\frac{i}{2},(1+i)x)}\\ y(x)\to \frac{(1+i)\mathrm{ParabolicCylinderD}(\frac{1}{2}+\frac{i}{2},(1+i)x)}{\mathrm{ParabolicCylinderD}(-\frac{1}{2}+\frac{i}{2},(1+i)x)}-ix\end{array}$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x**2 - y(x)**2 + Derivative(y(x), x) + 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

TypeError : bad operand type for unary -: list