1.92 problem 90

Internal problem ID [7784]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 90
Date solved : Monday, October 21, 2024 at 04:20:31 PM
CAS classification : [_Riccati]

Solve

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

1.92.1 Solved as first order ode of type Riccati

Time used: 0.438 (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*}

Solution obtained is

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

Taking derivative gives

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

Doing change of constants, the solution becomes

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

Figure 170: Slope field plot
\(y^{\prime }-y^{2}-x -x^{2} = 0\)

1.92.2 Maple step by step solution

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

1.92.3 Maple trace

Methods for first order ODEs:
 

1.92.4 Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 128

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

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

1.92.5 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*}