2.1.85 problem 84

Internal problem ID [8223]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 84
Date solved : Tuesday, November 12, 2024 at 11:07:36 PM
CAS classification : [_Riccati]

Solve

\begin{align*} y^{\prime }&=x^{2}+y^{2}-1 \end{align*}

Solved as first order ode of type Riccati

Time used: 0.345 (sec)

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= x^{2}+y^{2}-1 \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = x^{2}+y^{2}-1 \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=x^{2}-1\), \(f_1(x)=0\) 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}-1 \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} u^{\prime \prime }\left (x \right )+\left (x^{2}-1\right ) u \left (x \right ) = 0 \end{align*}

Solution obtained is

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

Taking derivative gives

\[ u^{\prime }\left (x \right ) = \frac {2 i c_1 \left (\left (\frac {1}{2}-\frac {1}{4 x^{2}}\right ) \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{x^{2}}\right ) x +2 i c_2 \left (\left (\frac {1}{2}-\frac {1}{4 x^{2}}\right ) \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\frac {i \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{x^{2}}\right ) x}{\sqrt {x}}-\frac {c_1 \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+c_2 \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{2 x^{{3}/{2}}} \]

Doing change of constants, the solution becomes

\[ y = -\frac {\left (\frac {2 i c_5 \left (\left (\frac {1}{2}-\frac {1}{4 x^{2}}\right ) \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{x^{2}}\right ) x +2 i \left (\left (\frac {1}{2}-\frac {1}{4 x^{2}}\right ) \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\frac {i \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{x^{2}}\right ) x}{\sqrt {x}}-\frac {c_5 \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{2 x^{{3}/{2}}}\right ) \sqrt {x}}{c_5 \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )} \]

Figure 2.173: Slope field plot
\(y^{\prime } = x^{2}+y^{2}-1\)

Summary of solutions found

\begin{align*} y &= \frac {\left (-3-i\right ) c_5 \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+4 \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) c_5 \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{2 x \left (c_5 \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )\right )} \\ \end{align*}

Maple step by step solution

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

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+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 dsolve solution

Solving time : 0.003 (sec)
Leaf size : 85

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

\[ y = \frac {\left (-3-i\right ) \operatorname {WhittakerM}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+4 \operatorname {WhittakerW}\left (1+\frac {i}{4}, \frac {1}{4}, i x^{2}\right ) c_{1} +\left (-2 i x^{2}+i+1\right ) \operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\left (-2 i x^{2}+i+1\right ) c_{1} \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )}{2 x \left (c_{1} \operatorname {WhittakerW}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )+\operatorname {WhittakerM}\left (\frac {i}{4}, \frac {1}{4}, i x^{2}\right )\right )} \]

Mathematica DSolve solution

Solving time : 0.225 (sec)
Leaf size : 153

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

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