Problem 27

2.1.27 Problem 27

2.1.27.1 Solved using ﬁrst_order_ode_riccati
2.1.27.2 ✓Maple
2.1.27.3 ✓Mathematica
2.1.27.4 ✗Sympy

Internal problem ID [10285]
Book : First order enumerated odes
Section : section 1
Problem number : 27
Date solved : Monday, January 26, 2026 at 09:33:06 PM
CAS classiﬁcation : [_Riccati]

2.1.27.1 Solved using ﬁrst_order_ode_riccati

4.606 (sec)

Entering ﬁrst order ode riccati solver

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

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

\[ u^{\prime \prime }\left (x \right )+\sin \left (x \right ) u \left (x \right ) = 0 \]

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

The solution for \(u \left (x \right )\) is

\begin{equation} \tag{3} 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 ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} 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} \end{equation}

Substituting equations (3,4) into (1) results in

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

Doing change of constants, the above solution becomes

\[ y = -\frac {\frac {\operatorname {MathieuCPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2}+\frac {c_3 \operatorname {MathieuSPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2}}{\operatorname {MathieuC}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )+c_3 \operatorname {MathieuS}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {-c_3 \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_3 \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 )} \\ \end{align*}


Direction ﬁeld \(y^{\prime } = \sin \left (x \right )+y^{2}\)	Isoclines for \(y^{\prime } = \sin \left (x \right )+y^{2}\)

Summary of solutions found

2.1.27.2 ✓ Maple. Time used: 0.002 (sec). Leaf size: 59

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

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

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) = -sin(x)*y(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 @ Moebi\ 
us 
            -> Mathieu 
               -> Equivalence to the rational form of Mathieu ODE under a powe\ 
r @ 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 step by step

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

2.1.27.3 ✓ Mathematica. Time used: 0.107 (sec). Leaf size: 105

ode=D[y[x],x]==Sin[x]+y[x]^2; 
ic={}; 
DSolve[{ode,ic},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*}

2.1.27.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-y(x)**2 - sin(x) + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)

NotImplementedError : The given ODE -y(x)**2 - sin(x) + Derivative(y(x), x) cannot be solved by the

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0

classify_ode(ode,func=y(x)) 
 
('1st_power_series', 'lie_group')

[next] [prev] [prev-tail] [front] [up]