2.1.27 problem 27

Internal problem ID [8437]
Book : First order enumerated odes
Section : section 1
Problem number : 27
Date solved : Tuesday, November 12, 2024 at 11:09:20 PM
CAS classification : [_Riccati]

Solve

\begin{align*} y^{\prime }&=\sin \left (x \right )+y^{2} \end{align*}

Solved as first order ode of type Riccati

Time used: 0.453 (sec)

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \sin \left (x \right )+y^{2} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=\sin \left (x \right )\), \(f_1(x)=0\) 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 &=\sin \left (x \right ) \end{align*}

Substituting the above terms back in equation (2) gives

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

Solution obtained is

\[ u \left (x \right ) = c_1 \operatorname {MathieuC}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )+c_2 \operatorname {MathieuS}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right ) \]

Taking derivative gives

\[ u^{\prime }\left (x \right ) = \frac {c_1 \operatorname {MathieuCPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2}+\frac {c_2 \operatorname {MathieuSPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2} \]

Doing change of constants, the solution becomes

\[ y = -\frac {\frac {c_1 \operatorname {MathieuCPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2}+\frac {\operatorname {MathieuSPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2}}{c_1 \operatorname {MathieuC}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )+\operatorname {MathieuS}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )} \]

Figure 2.28: Slope field plot
\(y^{\prime } = \sin \left (x \right )+y^{2}\)

Summary of solutions found

\begin{align*} y &= \frac {-c_1 \operatorname {MathieuCPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )-\operatorname {MathieuSPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2 c_1 \operatorname {MathieuC}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )+2 \operatorname {MathieuS}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )} \\ \end{align*}

Maple step by step solution

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

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_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -y(x)*sin(x), y(x)`      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
      -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
      -> Trying changes of variables to rationalize or make the ODE simpler 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Trying an equivalence, under non-integer power transformations, 
            to LODEs admitting Liouvillian solutions. 
            -> 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 
            -> Mathieu 
               -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
               Equivalence transformation and function parameters: {t = 1/2*t+1/2}, {kappa = -20, mu = -32} 
               <- Equivalence to the rational form of Mathieu ODE successful 
            <- Mathieu successful 
         <- special function solution successful 
         Change of variables used: 
            [x = arccos(t)] 
         Linear ODE actually solved: 
            (-t^2+1)^(1/2)*u(t)-t*diff(u(t),t)+(-t^2+1)*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful`
 

Maple dsolve solution

Solving time : 0.005 (sec)
Leaf size : 59

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

\[ y = \frac {-c_{1} \operatorname {MathieuSPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )-\operatorname {MathieuCPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2 c_{1} \operatorname {MathieuS}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )+2 \operatorname {MathieuC}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )} \]

Mathematica DSolve solution

Solving time : 0.174 (sec)
Leaf size : 105

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

\begin{align*} y(x)\to \frac {-\text {MathieuSPrime}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]+c_1 \text {MathieuCPrime}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]}{2 \left (\text {MathieuS}\left [0,-2,\frac {1}{4} (2 x-\pi )\right ]+c_1 \text {MathieuC}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]\right )} \\ y(x)\to \frac {\text {MathieuCPrime}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]}{2 \text {MathieuC}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]} \\ \end{align*}