2.1.92 problem 90

Internal problem ID [8480]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 90
Date solved : Tuesday, December 17, 2024 at 12:55:46 PM
CAS classification : [_Riccati]

Solve

\begin{align*} y^{\prime }-y^{2}-x -x^{2}&=0 \end{align*}

Solved as first order ode of type Riccati

Time used: 0.494 (sec)

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= x^{2}+y^{2}+x \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = x^{2}+y^{2}+x \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=x^{2}+x\), \(f_1(x)=0\) and \(f_2(x)=1\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \tag {1} \end{align*}

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{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=x^{2}+x \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} u^{\prime \prime }\left (x \right )+\left (x^{2}+x \right ) u \left (x \right ) = 0 \end{align*}

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

Solution obtained is

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

Taking derivative gives

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

Doing change of constants, the solution becomes

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

Figure 2.179: Slope field plot
\(y^{\prime }-y^{2}-x -x^{2} = 0\)

Summary of solutions found

\begin{align*} y &= -\frac {\left (2 \left (\operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )+\left (\frac {1}{24}+\frac {i}{2}\right ) \left (x +\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {7}{4}-\frac {i}{16}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right ) \left (2 x +1\right )+\left (\frac {1}{16}+\frac {i}{4}\right ) \operatorname {hypergeom}\left (\left [\frac {5}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right ) \left (2 x +1\right ) c_5 \right ) {\mathrm e}^{-\frac {i x \left (x +1\right )}{2}}+2 \left (\left (x +\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {1}{4}-\frac {i}{16}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right ) c_5}{2}\right ) \left (-\frac {i \left (x +1\right )}{2}-\frac {i x}{2}\right ) {\mathrm e}^{-\frac {i x \left (x +1\right )}{2}}\right ) {\mathrm e}^{\frac {i x \left (x +1\right )}{2}}}{2 \left (\left (x +\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {1}{4}-\frac {i}{16}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right ) c_5}{2}\right )} \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )-y \left (x \right )^{2}-x -x^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=y \left (x \right )^{2}+x +x^{2} \end {array} \]

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 \`\`^ @ Moebius\`\` is resolved 
         <- hypergeometric successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful`
 

Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 128

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

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

Mathematica DSolve solution

Solving time : 0.282 (sec)
Leaf size : 298

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

\begin{align*} y(x)\to \frac {i \left ((2 x+1) \operatorname {ParabolicCylinderD}\left (-\frac {1}{2}-\frac {i}{8},\left (-\frac {1}{2}+\frac {i}{2}\right ) (2 x+1)\right )-c_1 (2 x+1) \operatorname {ParabolicCylinderD}\left (-\frac {1}{2}+\frac {i}{8},(1+i) x+\left (\frac {1}{2}+\frac {i}{2}\right )\right )+(2+2 i) \left (\operatorname {ParabolicCylinderD}\left (\frac {1}{2}-\frac {i}{8},\left (-\frac {1}{2}+\frac {i}{2}\right ) (2 x+1)\right )-i c_1 \operatorname {ParabolicCylinderD}\left (\frac {1}{2}+\frac {i}{8},(1+i) x+\left (\frac {1}{2}+\frac {i}{2}\right )\right )\right )\right )}{2 \left (\operatorname {ParabolicCylinderD}\left (-\frac {1}{2}-\frac {i}{8},\left (-\frac {1}{2}+\frac {i}{2}\right ) (2 x+1)\right )+c_1 \operatorname {ParabolicCylinderD}\left (-\frac {1}{2}+\frac {i}{8},(1+i) x+\left (\frac {1}{2}+\frac {i}{2}\right )\right )\right )} \\ y(x)\to \frac {(1+i) \operatorname {ParabolicCylinderD}\left (\frac {1}{2}+\frac {i}{8},(1+i) x+\left (\frac {1}{2}+\frac {i}{2}\right )\right )}{\operatorname {ParabolicCylinderD}\left (-\frac {1}{2}+\frac {i}{8},(1+i) x+\left (\frac {1}{2}+\frac {i}{2}\right )\right )}-\frac {1}{2} i (2 x+1) \\ y(x)\to \frac {(1+i) \operatorname {ParabolicCylinderD}\left (\frac {1}{2}+\frac {i}{8},(1+i) x+\left (\frac {1}{2}+\frac {i}{2}\right )\right )}{\operatorname {ParabolicCylinderD}\left (-\frac {1}{2}+\frac {i}{8},(1+i) x+\left (\frac {1}{2}+\frac {i}{2}\right )\right )}-\frac {1}{2} i (2 x+1) \\ \end{align*}