2.1.92 Problem 90

Internal problem ID [8803]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 90
Date solved : Friday, April 25, 2025 at 05:11:43 PM
CAS classification : [_Riccati]

Solved using first_order_ode_riccati

Time used: 0.782 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }-y^2-x-x^2& =0\end{array}$$

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

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

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

Substituting the above terms back in equation (2) gives

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

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

Solution obtained is

$$u\left(x\right)=2(c_2(x+\frac{1}{2})\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+\frac{c_1\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4})}{2}){\mathrm{e}}^{-\frac{ix(x+1)}{2}}$$

Taking derivative gives

$$u^{\prime }\left(x\right)=2(c_2\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+(\frac{1}{24}+\frac{i}{2})c_2(x+\frac{1}{2})\mathrm{hypergeom}([\frac{7}{4}-\frac{i}{16}],\left[\frac{5}{2}\right],\frac{i{(2x+1)}^2}{4})(2x+1)+(\frac{1}{16}+\frac{i}{4})c_1\mathrm{hypergeom}([\frac{5}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})(2x+1)){\mathrm{e}}^{-\frac{ix(x+1)}{2}}+2(c_2(x+\frac{1}{2})\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+\frac{c_1\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4})}{2})(-\frac{i(x+1)}{2}-\frac{ix}{2}){\mathrm{e}}^{-\frac{ix(x+1)}{2}}$$

Doing change of constants, the solution becomes

$$y=-\frac{(2(\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+(\frac{1}{24}+\frac{i}{2})(x+\frac{1}{2})\mathrm{hypergeom}([\frac{7}{4}-\frac{i}{16}],\left[\frac{5}{2}\right],\frac{i{(2x+1)}^2}{4})(2x+1)+(\frac{1}{16}+\frac{i}{4})c_5\mathrm{hypergeom}([\frac{5}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})(2x+1)){\mathrm{e}}^{-\frac{ix(x+1)}{2}}+2((x+\frac{1}{2})\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+\frac{c_5\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4})}{2})(-\frac{i(x+1)}{2}-\frac{ix}{2}){\mathrm{e}}^{-\frac{ix(x+1)}{2}}){\mathrm{e}}^{\frac{ix(x+1)}{2}}}{2((x+\frac{1}{2})\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+\frac{c_5\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4})}{2})}$$

Which simplifies to

$$\begin{array}{rl}y& =\frac{(4i{x}^2+4ix+i-4)\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+4((-\frac{1}{12}-i)(x+\frac{1}{2})\mathrm{hypergeom}([\frac{7}{4}-\frac{i}{16}],\left[\frac{5}{2}\right],\frac{i{(2x+1)}^2}{4})+\frac{c_5((-\frac{1}{4}-i)\mathrm{hypergeom}([\frac{5}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+i\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4}))}{2})(x+\frac{1}{2})}{2\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})(2x+1)+2c_5\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4})}\end{array}$$

Figure 2.156: Slope field $y^{\prime }-y^2-x-x^2=0$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{(4i{x}^2+4ix+i-4)\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+4((-\frac{1}{12}-i)(x+\frac{1}{2})\mathrm{hypergeom}([\frac{7}{4}-\frac{i}{16}],\left[\frac{5}{2}\right],\frac{i{(2x+1)}^2}{4})+\frac{c_5((-\frac{1}{4}-i)\mathrm{hypergeom}([\frac{5}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+i\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4}))}{2})(x+\frac{1}{2})}{2\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})(2x+1)+2c_5\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4})}\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 128

ode:=diff(y(x),x)-y(x)^2-x-x^2 = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{2(i{x}^2+ix-1+\frac{1}{4}i)c_1\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+2((-\frac{1}{12}-i)(x+\frac{1}{2})c_1\mathrm{hypergeom}([\frac{7}{4}-\frac{i}{16}],\left[\frac{5}{2}\right],\frac{i{(2x+1)}^2}{4})+(-\frac{1}{8}-\frac{i}{2})\mathrm{hypergeom}([\frac{5}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+\frac{i\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4})}{2})(x+\frac{1}{2})}{(2x+1)c_1\mathrm{hypergeom}([\frac{3}{4}-\frac{i}{16}],\left[\frac{3}{2}\right],\frac{i{(2x+1)}^2}{4})+\mathrm{hypergeom}([\frac{1}{4}-\frac{i}{16}],\left[\frac{1}{2}\right],\frac{i{(2x+1)}^2}{4})}$$

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-x)*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 
         -> hypergeometric 
            -> heuristic approach 
            -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
            <- hyper3 successful: indirect Equivalence to 0F1 under ``^ @ Moebi\ 
us`` is resolved 
         <- hypergeometric successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}y\left(x\right)-y{\left(x\right)}^2-x-x^2=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=y{\left(x\right)}^2+x+x^2\end{array}$$

✓ Mathematica. Time used: 0.275 (sec). Leaf size: 298

ode=D[y[x],x]-y[x]^2-x-x^2==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}y(x)\to \frac{i((2x+1)\mathrm{ParabolicCylinderD}(-\frac{1}{2}-\frac{i}{8},(-\frac{1}{2}+\frac{i}{2})(2x+1))-c_1(2x+1)\mathrm{ParabolicCylinderD}(-\frac{1}{2}+\frac{i}{8},(1+i)x+(\frac{1}{2}+\frac{i}{2}))+(2+2i)(\mathrm{ParabolicCylinderD}(\frac{1}{2}-\frac{i}{8},(-\frac{1}{2}+\frac{i}{2})(2x+1))-i{c}_1\mathrm{ParabolicCylinderD}(\frac{1}{2}+\frac{i}{8},(1+i)x+(\frac{1}{2}+\frac{i}{2}))))}{2(\mathrm{ParabolicCylinderD}(-\frac{1}{2}-\frac{i}{8},(-\frac{1}{2}+\frac{i}{2})(2x+1))+c_1\mathrm{ParabolicCylinderD}(-\frac{1}{2}+\frac{i}{8},(1+i)x+(\frac{1}{2}+\frac{i}{2})))}\\ y(x)\to \frac{(1+i)\mathrm{ParabolicCylinderD}(\frac{1}{2}+\frac{i}{8},(1+i)x+(\frac{1}{2}+\frac{i}{2}))}{\mathrm{ParabolicCylinderD}(-\frac{1}{2}+\frac{i}{8},(1+i)x+(\frac{1}{2}+\frac{i}{2}))}-\frac{1}{2}i(2x+1)\\ y(x)\to \frac{(1+i)\mathrm{ParabolicCylinderD}(\frac{1}{2}+\frac{i}{8},(1+i)x+(\frac{1}{2}+\frac{i}{2}))}{\mathrm{ParabolicCylinderD}(-\frac{1}{2}+\frac{i}{8},(1+i)x+(\frac{1}{2}+\frac{i}{2}))}-\frac{1}{2}i(2x+1)\end{array}$$

✗ Sympy

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

TypeError : bad operand type for unary -: list