2.1.92 problem 90
Internal
problem
ID
[8230]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
90
Date
solved
:
Tuesday, November 12, 2024 at 11:07:38 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: 0.441 (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
\[
u \left (x \right ) = 2 \left (c_2 \left (x +\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )+\frac {c_1 \operatorname {hypergeom}\left (\left [\frac {1}{4}-\frac {i}{16}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )}{2}\right ) {\mathrm e}^{-\frac {i x \left (x +1\right )}{2}}
\]
Taking derivative gives
\[
u^{\prime }\left (x \right ) = 2 \left (c_2 \operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )+\left (\frac {1}{24}+\frac {i}{2}\right ) c_2 \left (x +\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {7}{4}-\frac {i}{16}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right ) \left (2 x +1\right )+\left (\frac {1}{16}+\frac {i}{4}\right ) c_1 \operatorname {hypergeom}\left (\left [\frac {5}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right ) \left (2 x +1\right )\right ) {\mathrm e}^{-\frac {i x \left (x +1\right )}{2}}+2 \left (c_2 \left (x +\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )+\frac {c_1 \operatorname {hypergeom}\left (\left [\frac {1}{4}-\frac {i}{16}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )}{2}\right ) \left (-\frac {i \left (x +1\right )}{2}-\frac {i x}{2}\right ) {\mathrm e}^{-\frac {i x \left (x +1\right )}{2}}
\]
Doing change of constants, the solution
becomes
\[
y = -\frac {\left (2 \left (\operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )+\left (\frac {1}{24}+\frac {i}{2}\right ) \left (x +\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {7}{4}-\frac {i}{16}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right ) \left (2 x +1\right )+\left (\frac {1}{16}+\frac {i}{4}\right ) c_5 \operatorname {hypergeom}\left (\left [\frac {5}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right ) \left (2 x +1\right )\right ) {\mathrm e}^{-\frac {i x \left (x +1\right )}{2}}+2 \left (\left (x +\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )+\frac {c_5 \operatorname {hypergeom}\left (\left [\frac {1}{4}-\frac {i}{16}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )}{2}\right ) \left (-\frac {i \left (x +1\right )}{2}-\frac {i x}{2}\right ) {\mathrm e}^{-\frac {i x \left (x +1\right )}{2}}\right ) {\mathrm e}^{\frac {i x \left (x +1\right )}{2}}}{2 \left (\left (x +\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {i}{16}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )+\frac {c_5 \operatorname {hypergeom}\left (\left [\frac {1}{4}-\frac {i}{16}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 x +1\right )^{2}}{4}\right )}{2}\right )}
\]
Figure 2.180: Slope field plot
\(y^{\prime }-y^{2}-x -x^{2} = 0\)
Summary of solutions found
\begin{align*}
y &= \frac {\left (4 i x^{2}+4 i x +i-4\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 )+4 \left (\left (-\frac {1}{12}-i\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 )+\frac {c_5 \left (\left (-\frac {1}{4}-i\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 )+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 )\right )}{2}\right ) \left (x +\frac {1}{2}\right )}{2 \left (2 x +1\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 c_5 \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 )} \\
\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*}